College Algebra with Current Interesting Applications and Facts by Acosta & Karwowski, Kendall Hunt, 2012

Textbook Supplement (Revised 8/2013)

Contents

A. Complex Numbers 2

B. Distance Formula, Midpoint Formula, and Circles 9

C. Synthetic Division 23

Topic Expansion from Content Already in Textbook

D. More on Rational Functions 33

E. Non-Linear Systems of Equations 47

F. Other Types of Equations 61

G. Solving Polynomial and Rational Inequalities 71

2

Section A: Complex Numbers

Suppose we wish to solve the equation x2 = −25. Since we know that the square of a real number is either zero or positive, there is no real number that would satisfy this equation. To solve this problem, mathematicians created a number system that is based upon a new number: the imaginary unit, commonly referred to as "i." Little Facts: It is said that the name "imaginary number" was originally coined by René Descartes in the seventeenth century as a derogatory term, because at that time such numbers were regarded by some as fictitious or useless. Swiss mathematician Leonhard Euler introduced the letter "i" to represent the square roots of negatives in 1777. In modern times these numbers have essential, concrete applications in math, physics, electrical engineering, and many other scientific and related areas. Sources: www.docstoc.com;www.nctm.org

The Imaginary Unit

The imaginary unit, i, is defined with the following properties

i = √−1 and i2 = √−1 • √−1 = −1

We can use the imaginary unit to rewrite the square root of a negative number.

Rewriting the Expression √−�

If a > 0, then √−� = i√�

Let us rewrite √−25 as a product of a real number and i: √−25 = �(−1)(25) = i√25 = 5i

We can check the answer by squaring 5i: (5i)2 = 52i2 = (25)(−1) = −25

Example 1

Rewrite √−28 in terms of i, and simplify if possible.

Solution

√−28 = i√28 = i√4•7 = 2i√7

Note: It is also acceptable to write an expression like 2i√7 as 2√7i, but we must be sure to write the "i" outside the radical symbol. To avoid being read as being under the radical, we generally write the answer with "i" in front of the radical.

When mathematicians added a real number to multiples of imaginary units, the set of complex numbers was formed.

Complex Number

A complex number is one of the form a + bi, where a and b are real numbers.

�In a complex number, we call a the real part and b is the imaginary part.

�Two complex numbers a + bi and c + di are equal if and only if a = c and b = d.

Any real number, a, can be written as a complex number as a + 0i. In this case, b = 0. If a complex number has b ≠ 0, then we call a + bi, an imaginary number (nonreal complex

3

number). On the other hand, if b ≠ 0 but a = 0, then a + bi = 0 + bi = bi and we call this a pure imaginary number. Some examples of complex numbers are:

2 + 5i imaginary number, a ≠ 0 b ≠ 0

7i pure imaginary number, a = 0

−12 real number, b = 0

A complex number written in the form a + bi is said to be in standard form. We have seen that the complex number system includes the set of real numbers as well as the set of imaginary numbers. The following diagram displays the relationship amongst complex, real, and imaginary numbers, where a and b are real numbers. Some examples are included.

Operations with Complex Numbers All the real properties of real numbers will hold for the complex numbers. We can add, subtract, multiply, and divide complex numbers using the real number properties and the operational definitions that we will discuss in this section.

Addition and Subtraction of Complex Numbers

For numbers a + bi and c + di, add or subtract the real parts and add or subtract the imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) − (c + di) = (a − c) + (b − d)i

Example 2

Perform the indicated operation and simplify. Write your answer in standard form.

a. (2 + 5i) + (7 −3i) b. (−4 − 8i) − (6 + i) c. (3 − 5i) + (1 +3i)− (17 − 2i)

Solution

a. (2 + 5i) + (7 −3i) = (2 + 7) + [5 + (− 3)]i = 9 + 2i

Complex Numbers a + bi

2 + 5�, 3 − 4�, �� + �

� �,i√2, 7�, −12, 0, �� , √3, π, √49, 8

Real Numbers a + bi, b = 0

−12, 0, �� , √3, π, √49, 8

Imaginary Numbers a + bi, b ≠ 0

2 + 5�, 3 − 4�, �� + �� �, �√2, 7�

4

b. (−4 − 8i) − (6 + i) = (−4 −6) + (− 8 − 1)i = −10 − 9i

c. (3 − 5i) + (1 +3i)− (17 − 2i) = (3 + 1 − 17) + (−5 + 3 + 2)i = −13 or −13 + 0i

Multiplication of Complex Numbers

For complex numbers a + bi and c + di, follow the same rules for multiplying binomials, remembering that i2 = −1.

(a + bi)(c + di) = ac + adi + bci + bdi2

= ac + adi + bci + bd(−1)

= (ac − bd) + (ad + bc)i

The good news is that we do not have to memorize this definition. We simply multiply the given complex numbers as with binomials.

Note: To simplify expressions containing complex numbers, start by rewriting any square roots with negative radicands as pure imaginary numbers. Example 3

Perform the indicated operation and simplify.

a. (2 − 5i)(6 +3i) b. √−16 • √−9 c. (3 + 7i)2 d. (3 + 5i)(3 −5i)

Solution

a. (2 − 5i) (6 +3i) = 12 + 6i −30i − 15i2 = 12 −24i − 15(−1) = 27 −24i

b. √−16 • √−9 = 4i • 3i = 12i2 = 12(−1) = −12

! CAUTION

Recall from elementary algebra that √� � √� = √�� only when both a and b are non-negative, thus √−16 • √−9 ≠ √144. c. (3 + 7i)2 = 32 + 2(3)(7i) + (7i)2 Square of a binomial

= 9 + 42i + 49(−1) = −40 + 42i

d. (3 + 5i)(3 −5i)= 32 − (5i)2 Product of sum and difference of same two terms

= 9 − 25(−1) = 34

From example 3(d), notice that when we multiplied the given complex numbers we obtained a real number. This will occur when we multiply a pair of complex numbers of the form (a + bi)(a − bi), which we call complex conjugates.

5

Property of Two Complex Conjugates

For real numbers a and b, (a + bi)(a − bi) = a2 + b2 (a − bi)(a + bi) = a2 + b2

For z = � + ��, some textbooks use the expression �� = � + ���������� to denote the complex conjugate.

We will use the aforementioned property to divide complex numbers.

Division of Two Complex Numbers

To obtain the quotient of two complex numbers, multiply the numerator and denominator by the complex conjugate of the denominator and simplify.

� + �� + !� = #� + ��

+ !�$ # − !� − !�$

Example 4 Express each quotient in standard form a + bi.

a.�&�'�(' b.

�'

Solution

a. We need to eliminate the imaginary part from the denominator, so that we can express the quotient in standard form. Multiplying both numerator and denominator by the complex conjugate, we will obtain a real number as the new denominator.�&�'

�(' = )�&�'�(' * )�&'

�&'*=�&�'&�'&�'+�('+ =�&�'&�'&�((�)

�(((�) =(�&�'� =− �

� +�� �

b. �' = )�' * )('

('*= (�'('+= (�'

(((�)=(�'� =−9i or 0 −9i

Example 5

Divide and simplify.

a. √(�,√(� b.

-&√(�.�

Solution

a. We first rewrite the square roots with negative radicands as pure imaginary numbers, then divide and simplify.

√(�,√(� ='√�,'√� =/�,

� =√32=√16•2=4√2

6

b. -&√(�.

� = -&'√�.

� = = -&'√0•�

� = -&�'√�

� = -� +

�'√�� = 4 + i√5

Powers of i

From the definition of the imaginary unit, we already know that that i2 = −1. Consider the following cyclic pattern:

�� = � i5 = i4 • i = 1 • i = �

i2 = −1 i6 = i4 • i2 = 1 • (−1) = −1

i3 = i2 • i = (−1) • i = −� i7 = i4 • i3 = 1 • (−i) = −� i4 = i2 • i2 = (−1)(−1) = 1 i8 = i4 • i4 = 1 • 1 = 1, and so on.

Observe that the powers of i repeat every fourth power, and the result is always one of only four possibilities: �, −1, −�, or 1. We can use the following steps to simplify higher powers of i.

Simplifying powers of i for 12, n > 4

1. Divide the exponent n by 4, and check the remainder, r.

2. Replace �3 with �4 (replace the original power of i with the remainder R)

3. Use the cyclic pattern �� = �, i2 = −1, i3 = −�, i4 =1 as needed to simplify.

Note: Recall that i0 =1 by the zero exponent property.

Example 6 Simplify each power of i.

a.i13 b.i30 c.�(�,

Solution

a.��0 =3, remainder R = 1. So, �� = i. Confirming the answer: i13 = i12 • i = (�0)3 • i =13 • i = � b.�.0 =7, R = 2. So, i2 = −1. Confirming the answer: i30 = (�0)7 • i2 = (1)7 • i2 =1 • ( −1) =−1

c. (�,0 =−4, R = 0. So, i0 = 1. Optionally, you can confirm the answer.

Graphing Calculator and Complex Numbers

Most graphing calculators have a complex numbers mode and we can use this as an optional tool for checking our work with complex numbers. We can use the M key to change the calculator from REAL numbers mode to a + bi mode. If we wanted to evaluate √−9 using the real numbers mode, the calculator will give us an error message, because this is not a real number. Using the complex number mode, the calculator will generate the correct answer, 3i. See the screens that follow.

7

The following screens show the answers we obtained in Examples 2a, and 2b, 3a and 3b, and 4a. You can confirm the other examples. In many calculators we can use ̀ . to type i.

! CAUTION

Remember to switch the calculator mode back to REAL numbers once you are finished working with complex numbers.

In-Class Practice Section (A) 1. Rewrite in terms of i, and simplify if possible:

a.√−81 b.√�13 c.√�200

2. Perform the indicated operation and simplify. Write your answer in standard form.

a. (5 + 2i) + (8 �i) b. (�10 � i) � (4 + 9i) c. (8 � 3i)(1 �14i)

d. √�100 • √�36 e. (3 � 5i)2 f. √(�..√(0 g.

,&'�(�'

h. �&√(5..

�. i. (√7 � i)(√7 �i) j. 4i(4 +5i) � (√�16 �6)

3. Simplify each power of i.

a.i19 b.i60 c.�(�-4. Confirm your answers to problem 2b and 2f with your graphing calculator.

8

Exercises Section (A)

In exercises 1-30, perform the indicated operation and simplify. Write answers in standard form.

1. (6 + 8i) + (9 �3i) 2. (�3 �20i) �(7 �13i) 3. (14 + 4i) −(10 −i) −17i

4. − (19 − 7i) +(−40 −2i) –(22 + 7i) 5. (2 − 5i)(3 +4i) 6. (−1 − 6i)(−9 −2i)

7. (−8 + i)(−4 −7i) 8. (9 + 7i)(−9 −7i) 9. (4 + 3i)2 10. (−3 − 7i)2

11. (−5 + 4i)(−5 −4i) 12. (6 − 9i)(6 + 9i) 13. (−2 −9i)2 14. (10 +�)2

15. √−49 • √−64 16. √−121 • √−4 17. −√−9 • √36 18. −√12 • √−3

19. 2

54− 20.

7

280

−−

21. 6

336

−−

22. 5

40− 23.

i

i

32

7

+−

24. i

i

26

3

+−

25. i

i

41

54

++

26. i

i

87

21

−−−

27. 3

726 −− 28.

2

528 −−

29. 5

1129 −+− 30.

9

754 −−−

In exercises 31-36, simplify each power of i.

31. 7i 32. 17i 33.

26i 34. 30i− 35.

59i− 36. 62i−

Answer Key Section (A) Exercises 1. 15 + 5� 3. 4 − 12� 5. 26 − 7� 7. 39 + 52� 9. 7 + 24� 11. 41

13. – 77 + 36� 15. – 56 17. – 18� 19. 33i 21. 142 23. i13

23

13

11−

25. i17

11

17

24 − 27. 222 i− 29. i5

74

5

9 +− 31. −� 32. � 33. −1 35. �

9

Section B: Distance Formula, Midpoint Formula, and Circles

Distance Formula

Suppose we want to determine the distance between the points 7�(3, 0) and 7�(8, 0) in the coordinate plane. By just looking at the graph, we can tell that these points are 5 units apart.

1 2 3 4 5 6 7 8 9

1

2

3

4

5

x

y

P1 P2

Points 7�and 7� have y-coordinate 0, and we know that they lie on the x-axis, a horizontal line. Recall that the distance between two points a and b on the number line is !�, � = |� � �|. So, the distance between two points on the x-axis is simply |9� � 9�|. Likewise, the distance between two points on the y-axis, a vertical line, is |:� � :�|. In our graph, the distance between 7�and 7� is |9� � 9�| = |8 − 3| = 5. Observe that we are not measuring the distance from point 7�to point 7�, which would imply direction. We are interested in finding the distance between the given points, regardless of direction. Thus, this distance can also be found by |9� − 9�| = |3 − 8| = |−5| = 5.

So, how do we calculate the distance between any 2 points on the coordinate plane, which we know will not always reflect horizontal or vertical lines? We can use the Pythagorean Theorem to develop a formula for the distance between any two points (9�, :�) and (9�, :�) in the coordinate plane. Let us construct a right triangle with a horizontal leg of length |9� − 9�| and a vertical leg of length |:� − :�|.

Since we have a right triangle, we can apply the Pythagorean Theorem, which states that the sum of the squares of its sides is equal to the square of the hypotenuse. Letting d represent the distance between (9�, :�) and (9�, :�), we have

!� = |9� − 9�|� + |:� − :�|�

We are squaring the quantities, thus we can replace the absolute value bars with parentheses.

!� = (9� − 9�)� + (:� − :�)�

x

y

(x1, y1)

(x2, y2)

(x2, y1)

|x2 - x1|

|y2 - y1|d

x1 x2

y1

y2

| |

_

_

10

Taking the principal square root, we now summarize the distance formula.

Distance Formula

The distance between points (9�, :�) and (9�, :�) in the coordinate plane is given by

! = �9� � 9�� � :� � :��

Example 1

Find the distance between the points (5, �7) and (�3, 4).

Solution

Using the distance formula, we have

! = �9� � 9�� � :� � :��

= ��3 − 5)� + [4 − (−7)]�

= �(−8)� + (11)�

= √64 + 121 = √185 ≈ 13.6

Hence, the exact distance between the given points is √185 units, or approximately 13.6 units.

Example 2

The circle below has diameter with endpoints at points P1 and P2. Calculate the diameter of this circle.

x

y

P1

P2 (1.2, 1.6)

(-1.2, -1.6)

Solution

Using the distance formula, we have

! = �[(1.2 − (−1.2)]� + [1.6 − (−1.6)]�

= �(2.4)� + (3.2)�

= √5.76 + 10.24 = √16 = 4

The diameter of the given circle has a length of 4 units.

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Midpoint Formula

As the name suggests, the midpoint of a line segment is the point that lies halfway between the endpoints of the line segment. If we have two points a and b on a number line, the midpoint, M, is located exactly midway between the two points, and we only need to average the coordinates, as shown below. = = >&?

�

Ma b

To find the coordinates =(9, : of the midpoint of the line segment that connects two points (9�, :�) and (9�, :�) on the coordinate plane, the formula will be similar, but now we must average both coordinates.

Since = lies exactly halfway between (9�, :�) and (9�, :�), we know that the distance between x1 and x and the distance between x and x2 is the same. Therefore, we know that x1 � x = x � x2. Solving for x, we have

x1 � x = x � x2 2x = x1 + x2 9 = @A&@+�

In the same way, : = BA&B+� . So, the location halfway between x1 and x2 is9 = @A&@+

� , and

between y1 and y2 is : = BA&B+� .

Midpoint Formula

The midpoint = = (9, : of the line segment with endpoints (9�, :�) and (9�, :�) in the coordinate plane is given by

= = #9� � 9�2 , :� + :�

2 $

Take the average of the two x-coordinates, and the average of the two y-coordinates.

! CAUTION

These are two things to keep in mind: • When using the distance formula, we subtract the respective coordinates; since we are

finding a distance, the answer is number. • When using the midpoint formula, we add the respective coordinates; since we are looking

for coordinates of a point, the answer is an ordered pair.

x

y

(x1, y1)

(x2, y2)M(x, y)

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Example 3

a. Find the midpoint of the line segment with endpoints (2, �7) and (�6, 5). b. Prove that the distance between (2, �7) and the midpoint equals the distance between (�6, 5) and the midpoint. c. Find half the distance between both endpoints.

Solution

a. Using the midpoint formula, we have

)�&(,� , (5&�

� * = )(0� , (�

� * = (�2, �1)

b. Distance between (2, �7) and (�2, �1) =��2 − 2)� + [−1 − (−7)]�

= √16 + 36 = √52 = √4•13 = 2 √13

Distance between (−6, 5) and (−2, −1) =�[(−2 − (−6)]� + (−1 − 5)�

= √16 + 36 = 2 √13 Both distances are equal.

c. �� [Distance between (2, −7) and (−6, 5)] = ���[(5 − (−7)]� + (−6 − 2)�

= ��√144 + 64

= ��√208 =

��√16•13 =

�� •4√13 = 2 √13

Observe that the distance between the midpoint and each of the endpoints is equivalent to half the distance between both endpoints.

Example 4

The graph below shows the approximate total revenue (in millions) for Google between the years 2007 and 2011. Assuming that the company's performance followed a linear pattern, use the midpoint formula to estimate its revenue for 2009. Source: www.google.com

2007 2008 2009 2010 2011

5000100001500020000250003000035000400004500050000

Year

Google Total Revenue ($ millions)

16,594

37,905

13

Solution

Using the midpoint formula, we have

)�..5&�.��� , �,��0&�5�.�

� * = (2009, 27249.5)

The estimate revenue for 2009 is $27,249.5 (in millions). Note: The actual revenue reported by Google for 2009 was $23,651 (in millions); the example assumes a linear pattern, which not necessarily reflects the actual company's performance trend.

Circles

How do we define a circle? A circle is the set of all points (9, : in the coordinate plane that are equidistant from a fixed point, (ℎ, D, called the center. The fixed distance between the circle's center to any point (9, : is called the radius, E, where E > 0.

x

y

r

(h, k)

(x, y)

We can use the distance formula to find the standard form of the equation of a circle. Let (ℎ, D represent the coordinates of the center of a circle, and assume that (9, :is any point on the graph of the circle. The distance between the points ℎ, D and (9, : is equal to the radius, E. Applying the distance formula, we have

�9 � ℎ� � : � D� = E

9 � ℎ� � : � D� = E� Square both sides.

This gives us the standard form of the equation of a circle.

The Standard Form of the Equation of a Circle

� For ℎ, D, and E real numbers, E> 0, the general form of the equation of the circle with center (ℎ, D and radius Eis given by

9 � ℎ� � : � D� = E�

� The equation of the circle with center at (0, 0) and radius E is given by

9� + :� = E�

Example 5

Find the standard form of the equation of each circle, given its center and radius.

a. Center (5, −2); radius 3 b. Center (2, 0); radius √5. c. Center (0,0); radius 1

14

Solution

a. We will use 9 � ℎ� � : � D� = E� with (ℎ, D = (5, �2) and E= 3.

9 � 5� � [: � �2)]� = 3� (9 − 5)� + (: + 2)� = 9

b. Since (ℎ, D) = (2, 0) and E = √5, we have

(9 − 2)� + :� = (√5)� (9 − 2)� + :� = 5

c. The center is located at the origin, thus we will use 9� + :� = E�with E = 1.

9� + :� = 1� 9� + :� = 1

Note: The circle with center at (0, 0) and radius E= 1, 9� + :�= 1, is called the "unit circle" and it is used frequently in trigonometry.

Example 6

Find the standard form of the equation of a circle with center at (2, −3), which passes through the point (3, 1).

Solution

We know that the radius Eof the circle is the distance between (2, −3) and (3, 1).

E = �(3 − 2)� +[1 − (−3)]�

= �(1)� + (4)� =√1 + 16 = √17 Since (ℎ, D) = (2, −3) and r = √17, the equation of the circle is

(9 − 2)� + [: − (−3)]� = F√17G�

(9 − 2)� + (: + 3)� = 17

Graphing Circles

Given the standard form of the equation of a circle, we can identify the center and the radius to help us graph the circle. Once we plot the center point, we count E units left and right from the center, and up and down from the center to plot four key points. Then, we connect the resulting points with a smooth curve to sketch the circle. The steps to graph the standard form of the equation of a circle are summarized next.

Graphing a Circle of the Form (H − I)J + (K − L)J = MJ

(1) Identify the center (ℎ, D) and the radius E.

(2) Plot the center point.

(3) Starting from (ℎ, D), count E units horizontally (left and right from the center) and vertically (up and down from the center) to plot four key points.

(3) Connect the four key points with a smooth curve to sketch the circle.

(4) Label the center and the radius.

15

Example 7 a. Graph the circle represented by 9 � 5� � : � 2)� = 9. Label the center and the radius. b. Find the intercepts, if any, of the graph of this circle.

Solution

a. The standard form of the equation of a circle gives its center and radius. Rewriting the given equation in the form (9 − 5)� + [: − (−2)]� = 3�, we know that the center (ℎ, D) = (5, −2), and the radius E= 3. Plotting the center, counting 3 units vertically and horizontally, and completing the circular sketch, we obtain the graph that follows.

x

y

r = 3

(5, -2)

(x, y)

b. To find any 9-intercepts, we let y = 0 and solve for x:

(9 − 5)� + (0 + 2)� = 9

(9 − 5)� + 4 = 9

(9 − 5)�= 5

9 − 5 = ±√5 9 = 5 ±√5

The x-intercepts are 9 = 5 +√5 ≈ 7.23 and 9 = 5 −√5 ≈ 2.76

c. To find any y-intercepts, we let x = 0 and solve for y:

(0 − 5)� + (: + 2)� = 9

25 + (: + 2)� = 9

(: + 3)� = −16

Since (: + 3)� cannot be negative, this circle does not have any :-intercept, as shown next.

9-intercepts

x

y

(2.76, 0) (7.23, 0)

16

General Form of the Equation of a Circle

Suppose we have the circle 9 � 3)� + (: + 1)� = 25. Expanding this equation, we have

9� − 69 + 9+:� + 2: + 1 = 25 Square each binomial.

9� − 69 + 9+:� + 2: + 1 − 25 = 0 Subtract 25 from both sides.

9�+:� − 69 + 2: − 15 = 0 Combine constants and rearrange terms.

The resulting equation is the equation of the circle written in general form.

The General Form of the Equation of a Circle

For A, B, C, D, and E real numbers, A = B, A and B not zero, the general form of the equation of the circle is given by

P9�+Q:� + R9 + S: + T = 0

Notice that while the standard form of the equation of the circle explicitly gives us information about its center and radius, the general form does not. When given the general form of the equation of a circle, we can rewrite the equation in standard form by using the process of completing the square. Example 8 Write each equation in standard form. Find the center and radius of each circle.

a. 9�+:� − 29 − 10: − 10 = 0 b. 49�+4:� − 49 + 4: = 1 Solution

a. We start by grouping any expressions involving the 9 variable, those involving the : variable, and adding 10 to both sides. 9�+:� − 29 − 10: − 10 = 0

(9�−29) + (:� − 10:) = 10

Next, we need to complete the square on both 9 and :. To complete the square on 9, square half the coefficient of 9 and add this square to both sides of the equation. Similarly, to complete the square on :, square half the coefficient of : and add this square to both sides of the equation. (9�−29 + 1) + (:� − 10: + 25) = 10 + 1 + 25 U�� (−2)V� = 1 U�� (−10)V� = 25

Add 1 to both sides Add 25 to both sides

Now we factor each trinomial, add the constants on the right-hand side of the equation, and write the equation in standard form.

(9 − 1)�+(: − 5)� = 36, or

(9 − 1)�+(: − 5)� = 6�

From the standard form of the equation we can see that the center of this circle is (1, 5) and its radius is 6.

17

b. To complete the square, the coefficients of 9� and:� must be 1.Therefore, dividing both sides of the equation 49��4:� � 49 � 4: = 1by 4, we obtain

9�+:� − 9 + : = �0

We can now rearrange and regroup the terms, and complete the square on both 9 and :.

(9� − 9)+(:� + :) = �0 Rearrange and regroup terms.

)9� − 9 + �0* + ):� + : + �

0* = �0 + �

0 + �0 Complete the square on both x and y.

U�� (−1)V� = �0 U�� (1)V

� = �0

Add �0 to both sides Add

�0 to both sides

)9 −��*�+ ): +��*

�=

�0 Factor and add the constants.

)9 −��*�+ ): +��*

�= )√�� *

�

So, the center of this circle is )�� , − ��* and its radius is √�

� ≈0.87.

! CAUTION

Although we can write every equation of a circle in the general form, not every equation of this form will represent the equation of a circle. For example, 9�+:� − 29 − 10: + 27= 0 is equivalent to 9�+:� − 29 − 10: = −27. Completing the square will result in (9 − 1)�+(: − 5)� = −1, which gives us E< 0. Thus, there are no real numbers 9 and :that satisfy this equation, and it has no graph. Additionally, let us consider the equation 9�+:� − 29 − 10: = −26. Completing the square, we obtain (9 − 1)�+(: − 5)� = 0. Since E= 0, this equation would not represent a circle, but rather a single point.

Example 9

The London Eye Ferris wheel has a diameter of 135 meters. Passengers get on the capsules on the lowest point of the wheel from a platform 2 meters above ground level. Assume that we establish a coordinate plane with the 9-axis running across ground level, and the center of the wheel located on the :-axis. Find the equation in standard form for the circular wheel. Little Facts: The London Eye Ferris wheel is located on London's South Bank, across the Westminster Bridge. It opened to the public in 2000 and, as of 2012, it was considered one of the most visited attractions in London, with approximately 3.5 million visitors per year. Interestingly, there are 32 passenger capsules on the wheel, but for superstitious reasons they are numbered 1-33; there is no capsule #13. Sources: www.londoneye.net; www.bbc.co.uk; www.howstuffworks.com; www.inventorsabout.com; www.design-technology.info; www.bordingarea.com;www.visitbritainshop.com

Solution

The center of the wheel is located on the :-axis, which means that its 9-coordinate is 0. Since the diameter is 135 meters, we know that the radius is 135 ÷ 2 = 67.5 meters. Given that the lowest part of the wheel comes within 2 meters of the ground, this means that the :-coordinate of the center will be at 67.5 + 2 = 69.5 meters. Therefore, the center of the wheel has coordinates

18

(0, 69.5). Thus, the equation of the circular wheel is (9 � 0��: � 69.5� = 67.5�, or equivalently, 9��: � 69.5� = 4556.25. Graphing Calculator and Circles

To graph the equation of the circle using the grapher, we must first isolate the :. Let us graph the equation 9� � :� = 4. Solving for : in terms of 9 we have

:� = 4 �9�

�:� = N √4 �9� : = N √4 �9� We let Y1 = √4 �9� to graph one-half of the circle, and Y2 = � √4 �9� to graph the other half. Since the default viewing window of the grapher is not a square, it is best use the ZSquare window (#, option 5) so that the graph does not look distorted. The graph is shown next.

[�4.7, 4.7, 1] by [�3.1, 3.1, 1]

In Example 7, we sketched the graph of 9 � 5� � : � 2� = 9. Solving for : in terms of 9,

we let Y1 = �2 � �9 � 9 � 5�, and for the other half we let Y2 = �2 � �9 � 9 � 5�. The graph follows.

[�5, 11.45, 1] by [�8.475, 2.375, 1]

Note: When we graph circles using the calculator, many times the circles will not appear completely closed because of the resolution of the calculator.

In-Class Practice Section (B) 1. Find the distance between each pair of points: a. (�8, 5) and (�3, �11) b. (3.5, 2) and (5, �6.2)

2. Two contestants in a reality show have to travel an undetermined number of miles to an unknown destination. They are only given the following information: Starting at coordinates (145, 0), the first day they need to travel to coordinates (160, 80); the second day they will travel west to a resting point with x-coordinate 25. What was the total distance they traveled? Round your answer to 2 decimal places.

3. Find the midpoint of each line segment with the given endpoints.

a. (10, 8) and (6, 22) b. )�� , ��5* and )� �

0 ,�,* c. √7, � √2 and √7, √2

19

4. a. Find the standard form of the equation of a circle with center (2, �1) and radius 2. b. Graph the circle and label the center and the radius. c. Find the intercepts, if any, of the graph of this circle.

5. Suppose a merchant ship is located at coordinates (64, �22) and has a radar system with 360° range of 20 nautical miles. Write the equation of the circle that would model the range of the radar. Note: One nautical mile ≈1.15 miles.

6. Write the equation 9��:� � 89 � 6: − 24 = 0 in standard form and identify the center and radius of the circle.

Exercises Section (B) In exercises 1-6, find the distance between the given pair of points. Write your answers in radical form, then round to 2 decimal places.

1. (−6, 7) and (2, −11) 2. (−3, −2) and (−5, −4) 3. (−9, 8) and (−1, −2)

4. (5, −7) and (−4, −3) 5. )− �� , 2*and )−1, ��* 6. )−2,− �

�* and )− 50 , −2*

In exercises 7-12, find the midpoint of each line segment with the given endpoints. Where needed write answer(s) in fraction form.

7. (−4, 2) and (6, −9) 8. (10, −1) and (−6, −4) 9. (−9, 3) and (3, −5)

10. (−6, 11) and (−5, −7) 11.)− 5� , 4*and )2,− �

�* 12. )�, , −8*and )− 5�� , ��*

In exercises 13-18, find the standard form of the equation of a circle with the given center and radius.

13. Center (4, 3); radius 4 14. Center (7,−2); radius 3 15. Center (−5, 4); radius 6

16. Center (−1,−3); radius 3 17. Center (6, 0); radius 8 18. Center (4.5, −3.8 ); radius 4.7

In exercises 19-24, find the standard form of the equation of a circle with the given center and specified point.

19. Center (−2, 3); point (4, 5) 20. Center (2, 3); point (4, 6)

21. Center (2, 1); point (−2,−3) 22. Center (−3, 7); point (−4, 2)

23. Center (5,−3); point (1,−2) 24. Center (−6,−7); point (−4,−10)

In exercises 25-29, graph the following circles; label the center and the radius.

25. (9 − 1)� + (: + 3)� = 5 26. (9 + 3)� + (: − 5)� = 7 27. (9 + 4)� + (: − 7)� = 3

28. (9 − 6)� + (: + 1)� = 4 29. (9 + 2)� + (: + 4)� = 9

In exercises 30-35, do the following:

a. Write in standard form. b. State the center. c. State the radius.

30. 9�+:� − 49 − 8: − 5 = 0 31. 9�+:� + 69 − 12: = 7

20

32. 9��14: + :� + 29 = 3 33. :�−209 + 9� + 10: − 4 = 0

34. 29�+2:� − 109 + 2: − 4 = 0 35. 59�+5:� − 25: − 159 − 5 = 0

In exercises 36-41, find the standard form for the given circles.

36. 37.

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-7

-6

-5

-4

-3

-2

-1

1 xy

(3, -1)

(2, -4)

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

12345678

x

y

(-3, 6)(-1, 4)

38. 39.

-7 -6 -5 -4 -3 -2 -1 1 2 3-1

1

2

3

4

x

y

(-5, 1)

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13 14

-7-6-5-4-3-2-1

12

x

y

(3, -1)

(7, -3)

40. 41.

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-2-1

1234567

x

y

(-2, 3)

(-5, 5)

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

x

y

(3, 1)

(4, 3)

42. Lake Eola’s fountain in downtown Orlando, Florida, has a radius of 30 feet. The lake is well-known for its swans and swan boats that are popular to locals. The bottom of the lake to the center of the fountain’s base is 24 feet. Assume that we establish a coordinate plane with the 9-axis running across the bottom of the lake, and the center of the fountain located on the :-axis. Find the equation in standard form for the circular fountain. Little Facts: Lake Eola was originally a sinkhole, and Lake Eola Park was established in 1888. The original sinkhole is north of the fountain with a depth of 80 feet. The first fountain was built in 1912, costing $10,000; the second fountain was built in 1957at a cost approximately $350,000. Source: www.cityoforlando.net

43. Randy is a drag racing enthusiast and had to buy new back tires, costing $390 a piece. He measured the radius of the tire to be 15.25 inches, and the width measuring 5.35 inches.

21

Assume that we establish a coordinate plane with the 9-axis running across the surface of the tire, and the center of the tire located on the :-axis. Find the equation in standard form for the circular tire. Round all values to 2 decimal places where needed. Little Facts: With a mere 6.5 to 7.5 psi, top fuel dragster back tires grow approximately 8.5 inches in diameter when reaching a 325 mph speed run. The dragsters typically accelerate from 0 to 100 mph in approximately just 0.84 seconds. Sources: www.motortrend.com; www.summitracing.com

44. The graph below shows the approximate opening stock price of Jarden Corporation at the beginning of each month from January to November in 2012. Assuming that the company's opening stock price at the beginning of each month followed a linear pattern, use the midpoint formula to estimate the opening stock price in June 2012. Round all values to 2 decimal places where needed. Little Fact: Jarden Corporation sells a diverse line of products; their portfolio contains, to name a few, Rawlings, Crock-Pot, Coleman, Oster and Mr. Coffee. Source: www.jarden.com

5

10

15

20

25

30

35

40

month

stock price

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov

(1, 20.41)

(11, 33.89)

45. Danny leaves Akron, Ohio and starts driving to the Rock and Roll Hall of Fame and Museum in Cleveland. His friend Eduardo notices that their place of departure is located at (8, 14) on a map’s coordinate grid. They stop halfway for lunch in Parma, then head towards the museum, whose coordinates are (20, 8). Find the location of Parma on their map. Little Facts: Alan Freed was a disc jockey in Cleveland, Ohio in 1953, in New York he was known as “The King of Rock ‘n’ Rollers” with thousands of followers. As he rose to fame, he was considered the Pied Piper that promoted juvenile delinquency via rock ‘n’ roll. The Rock and Roll Hall of Fame built on the shore of Lake Erie in 1995, is situated in Cleveland because it is where many historians say the first rock ‘n’ roll concert took place in 1952, at the Cleveland Arena, featuring “The Moondog Coronation Ball” that was a stage dance with R&B stars. Sources: www.rockhall.com; www.alanfreed.com

46. Suppose you situate a rectangular coordinate system on a baseball field in which home plate is the origin, home plate to 1st base is the positive 9-axis and home plate to 3rd base is the positive :-axis. The distance between each base is 90 feet. Assume a left fielder’s coordinate on the field is (85, 230). If the third baseman’s coordinate is (6, 96), find the distance that the left fielder’s ball will travel to 3rd base. Round your answer to the nearest foot.

47. A golfer is on his second stroke; his first stroke landed 32 feet short and 9 feet to the left of the 4th hole. His second stroke landed 6 feet past the hole and 4 feet to the right. Assuming that the hole is situated at the origin and the ball’s path was linear, find the distance his golf ball traveled on the second stroke. Round your answer to the nearest foot.

22

Answer Key Section (B) Exercises

1. 7.19972 ≈ 3. 81.12412 ≈ 5. 42.16

73 ≈ 7. (1, �7/2)

9. (�3. �1) 11. (�1/6, 9/5) 13. 9 � 4� � : � 3)� = 16

15. (9 + 5)� + (: − 4)� = 6 17. (9 − 6)� + :� = 64 19. (9 + 2)� + (: − 3)� = 40

21. (9 − 2)� + (: − 1)� = 32 23. (9 − 5)� + (: + 3)� = 17

25. 27.

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

xy

6=r

(1, -3)

-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-1

1

2

3

4

5

6

7

8

9

x

y(-4, 7)

3=r

29.

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-7-6-5-4-3-2-1

12

x

y

(-2, -4)

r=3

31. a. (9 + 3)� + (: − 6)� = 52 b. (−3, 6) c. 132

33. a. (9 − 10)� + (: + 5)� = 129 b. (10, −5) c. 129

35. a. 4

38

2

5

2

322

=

−+

− yx b. (3/2, 5/2) c. 2

38

37. (9 + 3)� + (: − 6)� = 8 39. (9 − 7)� + (: + 3)� = 20

41. (9 − 3)� + (: − 1)� = 5 43. 9� + (: − 5.35)� = 232.56 45. (14, 11) 47. 40 ft.

23

Section C: Synthetic Division

Division of Polynomials

We know that we can use long division to check whether or not the divisor is a factor of the dividend. Let us review the long division process.

Quotient

Divisor 21

29414 Dividend

�28 14 �14 0 Remainder Is 14 a factor of 294? The remainder is 0, thus 14 is a factor of 294. Checking: (14)(21) = 294.

Now, is the binomial9 � 3 a factor of 9� + 29� − 149 − 3? We can use long division of polynomials to answer this question. Polynomial division is very similar to long division of numbers. A first step is to make sure that both the dividend (9� + 29� − 149 − 3) and the divisor(9 − 3) are polynomials written in descending order. Let us review the process of long division of polynomials.

9� + 59 + 1

31423 23 −−+− xxxx Divide 9� by 9. So, the first term of the quotient is 92.

−( 9� − 39�) 9�(9 − 3) = 9� − 39�. Subtract; bring down next term.

59� −149 Divide 59� by 9. So, the second term of the quotient is 59. −(59� − 159) 59(9 − 3) = 59� − 159. Subtract; bring down next term. 9 − 3 Divide 9 by 9. So, the last term of the quotient is 1. −(9 − 3) 1(9 − 3) = 9 − 3. Subtract. 0 Remainder Since the remainder is 0, 9 − 3 divides evenly into 9� + 29� − 149 − 3 , and we conclude that 9 − 3 is a factor of 9� + 29� − 149 − 3. The quotient, 9� + 59 + 1, will be the other factor of the dividend. We can check by multiplying: (9 − 3)(9� + 59 + 1) = 9� + 29� − 149 − 3.

Example 1 Divide 99� + 159� − 14 by 39� + 1.

Solution

Observe that both the dividend and the divisor are missing first-degree terms. In this case, we insert 0's for the missing 9terms. The long division follows.

24

39 + 5

140159103 232 −++++ xxxxx Divide 99� by 39�.

−(99� + 09� + 39) Multiply, subtract, and bring down next term.

159� −39 −14 Divide 159� by 39�. −(159� + 09 + 5) Multiply, subtract, and bring down next term. −39 − 19 Remainder

When the remainder is 0 or the divisor has a higher degree than the remainder, we stop the division process. So, the quotient for this division is 39 + 5 with a remainder of −39 − 19.

The solution to a long division problem can be written as YZ[ZY\]YYZ[Z^_` =quotient + ̀

\abZ]Y\`YZ[Z^_` .

In our example, �@c&��@+(�0

�@+&� = 39 + 5 + (�@(���@+&� .

Next, we summarize the long division process.

The Division Algorithm

Let d(9) and !(9) be polynomials with !(9)≠0, and degree of !(9) less than the degree of d(9). There exist unique polynomials e(9) and E(9) such that

d(9)!(9) = e(9)+ E(9)

!(9) or

d(9) = !(9)•e(9) + E(9)

where E(9) = 0, or its degree is less than the degree of !(9). By the division algorithm, we can express the solution to Example 1 as follows.

99� + 159� − 14 = (39� + 1)(39 + 5) + (−39 − 19)

Synthetic Division

Synthetic division is shortcut method of dividing polynomials that can be used when the divisor is a first-degree polynomial of the form9 − , where is a constant. Using synthetic division, the process of division is simplified by omitting all the variables, and using only the coefficients of both dividend and divisor. Of course, we still use 0s for missing powers.

Let us divide 29� − 139 + 15 by 9 − 5 using long division, and compare it to the synthetic division of the same polynomials.

Dividend Divisor Quotient Remainder

Dividend Divisor Quotient Remainder

25

Long Division Synthetic Division

For the synthetic division process, we start by omitting all the variables, and using only the coefficients of both dividend and divisor. The coefficients of the dividend are 2, �13, and 15. The divisor 9 � 5 is equivalent to9 � �5, hence = 5. Since the coefficient of 9 in the divisor 9 � will always be 1, we can omit it when we work with synthetic division. We draw a line below the coefficients, leaving room above the line.

151325 2 +−− xxx

5 | 2 �13 15

Once we have set up the synthetic division (removing the variables), we bring down the leading coefficient.

151325 2 +−− xxx

5 | 2 �13 15

2 Now we multiply the divisor, 5, by the 2 we brought down, and write this product in the second row, second column. Then, add the values in the second column: (�13) + (10).

29

151325 2 +−− xxx 5 | 2 −13 15

−(292 − 109) 10 −39 + 15 2 −3

Next, multiply the divisor 5 by the −3 on the bottom row in the second column, and write this product in the second row, third column. Then, add the values in the third column: (15) + (−15).

29 − 3

151325 2 +−− xxx 5 | 2 −13 15

−(292 − 109) 2 10 −15 −39 + 15 2 −3 0

−(39 + 15) 0

The last entry of the final row on the synthetic division will give us the remainder of the division process. The other entries of the final row, will give us the coefficients (in descending order) of the terms of the quotient. The quotient will be a polynomial whose degree is 1 less than the degree of the dividend. In our example, we have

2 −3 0

Coefficients of the quotient Remainder

So, our quotient is 29 − 3, and we have a remainder of 0.

26

Example 2 Use synthetic division to divide �59� �39� − 9 by 9 + 1.

Solution

First, we rewrite the dividend in descending order, and insert 09: 39� − 59� + 09 − 9

The coefficients of the dividend are 3, −5, 0, and −9. Since the divisor is 9 + 1, which is equivalent to 9 − (−1), the value of the constant is −1.

−1 | 3 −5 0 −9 Set up the synthetic division (remove the variables) and bring down the leading coefficient.

3 −1 | 3 −5 0 −9 Multiply the divisor, −1, by the 3 on the bottom row and write the

−3 product in the second row, second column. Add the values in the second column: (−5) + (−3). 3 −8

−1 | 3 −5 0 −9 Multiply −1 by −8 on the bottom row and write the product in the −3 8 second row, third column. Add the values in this column: 0 + 8.

3 −8 8

−1 | 3 −5 0 −9 Multiply −1 by 8 on the bottom row and write the product in the

−3 8 −8 second row, next column. Add these values: (−9) + (−8).

3 −8 8 −17

The entries 3, −8, and 8 will give us the coefficients (in descending order) of the terms of the quotient, and the remainder is the last entry, −17. So, dividing 39� − 59� − 9 by 9 + 1 results in

39� − 89 + 8 + (�5@&� .

It is important to remember that the degree of the quotient will be 1 less than the degree of the dividend. Next, is a summary of the synthetic division process.

27

Synthetic Division

To divide a polynomial d9 by a first-degree polynomial of the form 9 � , where is a constant, we can follow these steps.

(1) Write the coefficients of dividend, d9, in descending order, and insert 0's for any missing terms.

(2) Determine the value of (from the divisor) and set up the synthetic division.

(3) Draw a line below the coefficients, leaving room above the line, and bring down the leading coefficient.

(4) Multiply by the coefficient you brought down, and write this product in the second row, second column. Add the values in the second column.

(5) Multiply the divisor, , by the result from step 4, and write this product in the second row, third column. Add the values in the third column.

(6) Repeat the process until all the coefficients have been used and all columns have been filled in.

(7) The last entry of the bottom row is the remainder of the division process. The other entries in this row will give the coefficients (in descending order) of the terms of the quotient, which will be a polynomial whose degree is 1 less than the degree of d9. With practice, we can abbreviate the process. We will show the shorter version using the problem from Example 2. �1 | 3 �5 0 �9 Set up the synthetic division format.

�1 | 3 �5 0 �9 Bring down the leading coefficient, 3

�3 8 �8 (�1)(3) = �3; �5)��3) = �8

3 �8 8 �17 (�1)(�8) = 8; 0 + 8 = 8 (�1)(8) = �8; �9)��8) = �17

The quotient is 39� − 89 + 8 and the remainder is −17. The Remainder Theorem

Notice that the remainder obtained when we divide a polynomial function d(9) by a first-degree polynomial, 9 − , is always a constant. This occurs because the degree of the remainder must be less than 1 (that is, less than the degree of 9 − ).

Let us revisit the Division Algorithm, d(9) = !(9)•e(9) + E(9). Letting !(9) = 9 − , and knowing that the remainder E(9) will be a constant, we have

d(9) = (9 − )•e(9) + E

Dividend Divisor Quotient Remainder

28

Suppose we evaluate d9 when 9 = . We get

d = � •e � E

d = 0)•e( ) + E

d( ) = E

This leads us to a new way of evaluating polynomials, which we state next.

The Remainder Theorem

If a polynomial d(9) is divided by 9 − , then the remainder is d( ).

In Example 2 we used synthetic division to divide d(9) = 39� − 59� − 9 by 9 + 1 and the remainder was −17. Since the divisor is equivalent to 9 − (−1), the value of = −1. Let us now substitute −1 for 9 and evaluate d(−1). d(−1) = 3(−1)� − 5(−1)� − 9

d(−1) = −3 − 5 − 9 = −17

Observe that d(−1) = −17, which is the remainder we obtained via the synthetic division ofd(9) by 9 + 1.

The remainder theorem implies that we can use synthetic division to evaluate a polynomial function d(9) at 9 = . When we divide d(9) by 9 − the remainder will be d( ), that is, the value of the polynomial at .

Example 3 Let d(9) = 590−39� + 9� − 9 + 9. Use the remainder theorem to find d(−2). Solution

We will use synthetic division with = −2. The coefficients of the dividend are 5, −3, 1, −1, and 9.

−2 | 5 −3 1 −1 9 −10 26 −54 110

5 −13 27 −55 119 By the remainder theorem, d(−2) =119. Remainder Optionally, we can verify our finding by direct evaluation:

d(−2) = 5(−2)0−3(−2)� + (−2)� − (−2)+ 9 = 80 + 24 + 4 + 2 + 9 = 119

29

The Factor Theorem

Earlier, we divided 9� � 29� − 149 − 3 by 9 − 3 and we obtained the quotient 9� + 59 + 1 with a remainder of 0. Since 9 − 3 divides evenly into 9� + 29� − 149 − 3, we were able to conclude that 9 − 3 was a factor of the dividend. Checking our answer, we noticed that the multiplication of (9 − 3)(9� + 59 + 1) = 9� + 29� − 149 − 3. By the remainder theorem, if d(9) is divided by 9 − , and the remainder is 0, then we know that d( ) = 0. Next, we summarize the relationship between the first-degree factor, 9 − , the constant , and d( ) = 0.

Factor Theorem

If d(9) is a polynomial function, then9 − is a factor of d(9) if and only if d( ) = 0. This implies that if d( ) = 0 then9 − is a factor of d( ) = 0.

Note: The Factor Theorem is revisited in Section 7.1.

We can use the remainder theorem and the factor theorem to determine factors of polynomials of higher degrees. Example 4 Determine if 9 + 5 is a factor of the given polynomials.

a. d(9) = 9� + 29� − 139 + 10 b. d(9) = 490−9� + 29 + 4

Solution

a. 9 + 5 is equivalent to 9 − (−5), and the value of the constant is −5. By the factor theorem, if 9 − is a factor of d(9), thend( ) = d(−5) = 0. Using synthetic division, we have

−5 | 1 2 −13 10 −5 15 −10

1 −3 2 0

By the remainder theorem, d(−5) =0. Therefore, we know that 9 + 5 is a factor of d(9) = 9� + 29� − 139 + 10.

b. Using synthetic division,

−5 | 4 −1 0 2 4 −20 105 −525 2615

4 −21 105 −523 2619

By the remainder theorem, d(−5)= 2619. Since the remainder is not 0, we conclude that 9 + 5 is not a factor of (9) = 490−9� + 29 + 4.

Example 5

Given that 9 + 2 is a factor of d(9) = 29� + 9� − 59 + 2, find all possible real zeros.

Solution

Using synthetic division, we have the following:

30

�2 | 2 1 �5 2

�4 6 �2

2 �3 1 0

The quotient is 29� − 39 + 1 and the remainder is 0, hence we can write the given polynomial as (9 + 2)(29� − 39 + 1). We can factor the quotient as (29 − 1)(9 − 1). Thus, the factorization of d(9) = 29� + 9� − 59 + 2 is given byd(9) = (9 + 2)(29 − 1)(9 − 1). The zeros can be found when d(9) = (9 + 2)(29 − 1)(9 − 1) = 0. So, the real zeros of the given

polynomial are −2, �� and 1.

Note: if 9 + 2 is a factor of a polynomial, −2 is a zero. In general, if 9 + is a factor of a polynomial, − is a zero of that polynomial.

The attendance, P, (in millions) to Magic Kingdom, in Orlando, FL, for the years 2005 to 2011 can be modeled by P(9) = 0.019� − 0.159� + 0.649 + 16.17, where 9 represents the number of years since 2005. Comparing the years 2008 and 2010, which had a higher attendance? Use the remainder theorem to answer the question. Sources: www.themeparkinsider.com; www.coastergrotto.com

Solution Since 9 represents the number of years since 2005, 9 = 3 represents 2008, and 9 = 5 is 2010. First, we will find P(3). 3 | 0.01 −0.15 0.64 16.17

0.03 −0.36 0.84

0.01 −0.12 0.28 17.01 P(3) = 17.01 Next, we will find P(5).

5 | 0.01 −0.15 0.64 16.17

0.05 −0.5 0.7

0.01 −0.1 0.14 16.87 P(5) = 16.87

The attendance (in millions) in 2008 was 17.1, and in 2010 it was 16.87. Therefore, 2008 had more visitors. Note: The actual attendance (in millions) reported by Magic Kingdom for 2008 and 2010 was 17.0 and 16.9, respectively, which is very close to the approximation provided by our model.

In-Class Practice Section (C) 1. Use long division to divide: (169� + 129 − 7) ÷ (89 + 4).

2. Use synthetic division to divide: (−390 +9� − 99 + 5) ÷ (9 + 1). 3. Let d(9) = 690−9� + 59� − 9 − 3. Use the remainder theorem and synthetic division to find d(−3).

Example 6

31

4. Determine if 9 � 4 is a factor of the given polynomials. a. d9 = 290 − 69� − 79� − 69 − 3 b. d(9) = 9�−29� − 179 + 28

5. Given that 9 − 6 is a factor of d(9) = 9� − 49� − 159 + 18, find all possible real zeros.

6. The attendance, P, (in millions) to Disneyland, in Anaheim, CA, for the years 2005 to 2011 can be modeled by P(9) = −0.0199� + 0.29� − 0.2429 + 14.581, where 9 represents the number of years since 2005. Comparing the years 2007 and 2011, which had a higher attendance? Use the remainder theorem to answer the question. Sources: www.themeparkinsider.com; www.coastergrotto.com

Exercises Section (C) In exercises 1-6, divide using long division. State the quotient and the remainder.

1. )2()4022( 23 +÷−−+ xxxx 2. )6()3624( 23 −÷−−− xxxx 4. )12()224( 23 −÷−+− xxxx 4. ( 392512 3 +−− xxx )÷( 53 +x ) 5. )26()51218( 223 −÷−+− xxx 6. )32()54828( 223 −−÷−−+− xxxx In exercises 7-11, divide using synthetic division. State the quotient and the remainder.

7. )2()1342( 23 −÷−+− xxxx 8. )3()5494( 23 +÷+−−− xxxx 9. )1()127( 3 +÷+−− xxx 10. )4()93( 3 −÷−− xxx 11. )4()3022( 24 +÷−+− xxx In exercises 12-15, given the following polynomials use the remainder theorem and synthetic division to find the indicated function value.

12. 8273)( 234 −+−+−= xxxxxf ; )1(−f 13. 5247)( 32 +−+−= xxxxg ; )2(g

14. 7122524)( 32 −+−= xxxxh ; )6(h 15. 32 7158)( xxxf ++−= ; )2(−f 16. Determine if 3+x is a factor of the given polynomials.

a. 234 642)( xxxxf −+=

b. 2425)( 23 +−−= xxxxg

c. 92.136.86)( 23 −++= xxxxp In exercises 17-22, find all possible real zeros of each polynomial, given its factor. Write answers in fraction form where needed.

17. ;12783)( 23 −−+= xxxxp )1( +x 18. ;1756092)( 23 −−+= xxxxf )7( +x 19. ;2414116)( 23 +−−= xxxxf )2( −x 20. ;927348)( 23 −++= xxxxh )3( +x 21. ;18938016)( 23 −−−= xxxxg )6( −x 22. ;501258312)( 23 −+−= xxxxf )5( −x

32

23. The attendance, P, (in thousands) to a baseball pitching tournament for the years 1990 to 2010 can be modeled by P9 = 9� � 35.59� + 416.869 − 1617, where 9 represents the number of years since 1990. Comparing the years 2007 and 2008, which year had higher attendance? Round the attendance number to 2 decimal places. Use the remainder theorem to answer the question. Little Facts: The rotation of the knuckleball is vastly different from the other 3 main pitches, that is, the curveball, fastball, and slider. A typical fastball will make around 16 revolutions, whereas a typical knuckleball will make between 1.5 to 2.5. The reason a knuckleball is so difficult to pitch is that it is more pushed rather than thrown. Sources: www.newsfeed.time.com; www.popularmechancis.com; www.baseball.physics.illionois.edu

24. Sally owns a gummi bear distribution company that specializes in all natural flavored gummies. The sales of her company, f, (in thousands) for the years 2000 to 2012 can be modeled by f(9) = 2.39� − 4.19� + 2.259 + 4.2, where 9 represents the number of years since 2000. Comparing the years 2005 and 2011, which year had higher sales? Round the sales to 2 decimal places. Use the remainder theorem to answer the question. Little Facts: German Hans Riegel invented the first gummi bears in the 1920’s. He soon formed the company Haribo, which stands for Hans Riegel Bonn. The first Haribo gummi bear was sold in the US in 1982. Sources: www.investors.about.com; www.haribo.bg/en

Answer Key Section (C) Exercises

1. 202 −− xx 3. 1:;32 2 Rx + 5. 16:;23 −−+− xRx 7. 5:;32 2 Rx +

9. 10:;977 2 Rxx −+− 11. 30:;8282 23 Rxxx −+− 13. 5 15. −4

17. −3, 0� , −1 19. − �� , 0� , 2 21. − �

0 , − �0 , 6 23. 2008; 216.48 (in thousands)

33

Section D: More on Rational Functions

In section 7.2, we learned about vertical and horizontal asymptotes of rational functions. Let us revisit the aforementioned with an example.

Example 1

Given the rational function: d9 = @+(�@(�

a. Find any horizontal or vertical asymptotes. b. Graph the function along with any horizontal or vertical asymptote(s).

Solution

a. Function dis already in lowest terms. Horizontal asymptote: Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Vertical asymptote: Solving 9 � 2 = 0, we obtain 9 = 2.

b. We already determined the asymptote. Following the steps delineated in Section 7.2, we have

� The domain of the function is the set of all real numbers except 2, or (−∞,2∪(2,∞. � d(0 =

(.+(�.(� =

(�(� = 1, thus the y-intercept is (0, 1).

� Setting the numerator 9� − 2 = 0 and solving for x, we get x = ±√2. The x-intercepts or zeros for the given function are (−√2 0) ≈ (−1.4142, 0 and (−√2 0) ≈ (1.4142, 0.

� The following table of values contains additional points to helps us sketch the graph:

The graph is displayed below.

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-6-4-2

2468

101214

x

y

x = 2

22

)(2

−−=

xx

xf

Oblique (Slant) Asymptotes

Some graphs of rational functions have another type of asymptote, which is neither horizontal nor vertical. When the degree of the numerator is exactly one more than the degree of the denominator, an oblique asymptote occurs. This asymptote also gives us information about the behavior of the graph as x −∞ (that is, as x approaches negative infinity) and as x ∞ (as x approaches positive infinity).

x −6 −2 1 3 6

f(x) −4.25 0.5 1 7 8.5

34

Note: Some authors prefer to use the word slant rather than oblique. Both terms may be used synonymously. We will use the term oblique throughout this topic.

Let us consider the function from Example 1. To find an oblique asymptote, we divide the numerator, 9� � 2, by the denominator, 9 − 2. Since we have a linear divisor of the form 9 − , we can use synthetic division with = 2. The coefficients of the dividend are 1, 0, and −2.

2 | 1 0 −2 2 4

1 2 2

The quotient is 9 + 2 and the remainder is 2. Thus, d(9 = 9 + 2 + �

@(� .

Observe that as x −∞ or x ∞, the expression �

@(� approaches 0, and we see that the

graph of the function d approaches the graph of : = 9 � 2, So, we say that the line : = 9 + 2 is an oblique asymptote of the graph of d. The graph from Example 1, along with the vertical and oblique asymptotes, is shown below.

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-6-4-2

2468

101214

x

y

y = x + 2

x = 2

22

)(2

−−=

xx

xf

Oblique (Slant) Asymptote of a Rational Function

� Let d(9 = i(@j(@ be a rational function in lowest form, such that the degree of k(9 is

exactly one more than the degree of !(9. The graph of d will have an oblique asymptote.

� To find the oblique asymptote, divide k(9 by !(9, where the quotient is e(9 and the remainder is E(9

d(9 = e(9 + 4(@j(@

Since the degree of k(9 is one more than the degree of !(9, the quotient e(9 is a linear function, which will be the oblique asymptote of d. The remainder becomes almost zero, and we discard it.

Following is a summary of the different linear asymptotes of rational functions.

35

Summary of Linear Asymptotes of Rational Functions

Let d9 = i@j@ be a rational function such that k9 and !9 have no common factors.

Vertical Asymptotes: The line 9 = D is a vertical asymptote of graph of the function if !D = 0. A rational function may have many vertical asymptotes or none at all; the graph of dwill never intersect a vertical asymptote.

Horizontal Asymptotes: The line : = D is a horizontal asymptote of the graph of the function if d9 approaches Das 9 �∞or as 9 ∞.

Let d9 = i@j@ =

>l@l&>lmA@lmA&…&>+@+&>A@&>o?p@p&?pmA@pmA&…&?+@+&?A@&?o , where �q ≠ 0, �3 ≠ 0, r = degree of

the numerator, and s = degree of the denominator.

� If r< s, the line : = 0 (that is, the 9-axis) is the horizontal asymptote of d. � If r = s, the line :=

>l?p (ratio of the leading coefficients) is the horizontal asymptote of d.

� If r > s, the function does not have a horizontal asymptote.

Oblique Asymptotes: If the degree of k9 is exactly one more than the degree of !9, the graph of d will have an oblique asymptote.

� Divide the numerator by the denominator and discard the remainder. The quotient is a linear equation, : = r9 � �, which will be the oblique asymptote of d.

A rational function may have at most one horizontal or one oblique asymptote, but never both; the graph of d may intersect its horizontal or its oblique asymptote.

Note: Remember that an asymptote is not part of the graph of the function.

In section 7.2, we examined properties and graphs of rational functions, including any corresponding vertical and horizontal asymptotes, and removable discontinuities or holes. We also saw their applications in different real-world scenarios.

In this section, we have already amplified our knowledge of asymptotes, with rational functions that contain oblique asymptotes. We will now expand our method for graphing rational functions.

Next, is a summary of the strategies we can apply to sketch the graph of a rational function.

36

Strategy for Sketching the Graph of a Rational Function

To graph a given rational function of the form f(x) = i@t@, wherek9 and e9 are polynomials

and e9 ≠ 0, follow these guidelines:

(1) Factor k9 and e9, and determine the domain of the rational function; restrict any input values as needed.

(2) Write the rational function in lowest terms.

(3) Determine the intercepts of the graph: Find and plot the :-intercept, if it exists, by evaluating

d0. Find and plot any 9-intercepts (zeros) by solving the equation p(x) = 0.

(4) Find any vertical asymptotes. Sketch any existing vertical asymptotes using dashed lines. Check for holes (if any).

(5) Find the horizontal or oblique asymptote, if there is one. If a horizontal or oblique asymptote exists, sketch it using a dashed line. Determine any points where the graph of the rational function may intersect these asymptotes, if one of them exists. We can do this by equating the rational function and the asymptote and finding the solutions, if any.

(6) Plot enough points to determine the behavior of the graph of the function between and beyond any 9-intercepts and any vertical asymptotes.

(7) Check whether the graph of d has symmetry. The graph is symmetric about the :-axis if the function is even, d(−9= d(9. The graph is symmetric about the origin if the function is odd, d(−9= −d(9. (8) Draw the graph with a smooth curve.

Note: Although asymptotes are not part of the graph, they indicate the behavior of the function as it approaches them.

Example 2

Sketch the graph of d(9 = @+(,@(5@(� . Label any asymptotes.

Solution

(1) d(9 = @+(,@(5@(� = (@(5(@&�

@(� . The domain of the function is (−∞,1∪1,∞. (2) Function dis in lowest terms.

(3) d(0 = (.+(,(.(5

.(� = 7, thus the y-intercept is (0, 7). Solving 9� − 69 − 7 = 0, we have the

9-intercepts (−1, 0) and (7, 0).

(4) Solving 9 − 1 = 0, we obtain 9 = 1.The graph has a vertical asymptote at 9 = 1.

(5) Since the degree of the numerator is one more than the degree of the denominator, this function does not have a horizontal asymptote, and it has an oblique asymptote. Dividing

9� � 69 − 7 by 9 − 1, we have

1 | 1 �6 �7 1 �5

1 �5 �12

37

d9 = 9 � 5 � (��

@(� , therefore, : = 9 � 5 is an oblique asymptote of the graph of d.

We can equate the rational function and the asymptote, and solve, to determine if the graph of dwill intersect this asymptote.

@+(,@(5

@(� = 9 � 5

9� � 69 − 7 = 9� − 69 + 5 −7 ≠ 5

There is no solution; hence, the graph will not intersect the oblique asymptote.

(6) We determine some additional points to helps us sketch the graph.

(7) d(−9 = @+&,@(5(@(� . Since d(−9 does not equal either d(9 or −d(9, the graph of this

rational function is not symmetric about the :-axis nor the origin.

(8) The graph is shown next.

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

-20-18-16-14-12-10-8-6-4-2

2468

1012

x

y

y = x - 5

x = 11

76)(

2

−−−=

xxx

xf

When we graph rational functions, not all asymptotes will be linear. For example, let

d(9 = @u(�@c&�

@+ . Notice that the degree of the numerator is two more than the degree of

the denominator. The division of 90 − 29� + 1 by the denominator, 9�, will result in

d(9 = 9� − 29 + �@+, which tells us that we have a non-linear quotient. Thus, the asymptote of

this function will be : = 9� − 29. The graph, along with the function's vertical asymptote and non-linear asymptote, follows.

-2 -1 1 2

-5-4-3-2-1

1234567

x

y

2

34 12)(

x

xxxf

+−=

y = x2 - 2x

x = 0

x −3 −2 2 3 5

d(9 −5 −3 -15 -8 −3

38

Example 3

Sketch the graph of each function. Label any asymptote(s) and hole(s).

a. d9= @+&@(��@+(0 b. d9 =

@c&�@+(� c. d9 =

@u&�@+

Solution

a. (1) d9 = @+&@(��@+(0 =

@&0@(�@(�@&�. The domain is the set of all real numbers except �2

and 2, or (�∞,�2∪(−2,2∪(2,∞. (2) Function dis in lowest terms.

(3) d(0 = 3, thus the :-intercept is (0, 3). Solving 9� + 9 − 12 = 0, we get the 9-intercepts (−4, 0) and (3, 0).

(4) Solving 9� − 4 = 0, we obtain 9 = −2 and 9 =2. There will be vertical asymptotes at 9 = −2 and 9 =2

(5) Since the degree of the numerator is equal to the degree of the denominator, and the coefficient on both 9� terms is 1, there will be a horizontal asymptote at : = 1. Solving d(9 = 1 will tell us whether the graph of f intersects the horizontal asymptote.

d(9 = @+&@(��@+(0 = 1

9� + 9 − 12 = 9� − 4

9 − 12 = −4

9 = 8

Therefore, the graph of f intersects the horizontal asymptote (line : = 1) at (8, 1).

(6) We determine some additional points to helps us sketch the graph. The table feature of the calculator is a useful tool for this step.

(7) d(−9 = @+(@(��@+(0 . Since d(−9 does not equal either d(9 or −d(9, the graph of d is

not symmetric about the :-axis nor the origin.

(8) The graph is shown next.

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

-12

-8

-4

4

8

x

y

y = 1

x = 2

4

12)(

2

2

−−−=

x

xxxf

x = -2

x −3 −2.5 −1 0.5 1.5 2.5

d(9 −1.2 −3.7 4 3 4.7 1.4

39

b. (1) d9 = @c&�@+(� =

@&�@+(@&�@&�@(� . The domain is the set of all real numbers except �1

and 1, or (�∞,�1∪(−1,1∪(1,∞.

(2) Functiondin lowest terms is: d(9 = @+(@&�

@(� .

(3) d(0= −1; the :-intercept is (0, −1). The equation 9� − 9 + 1 = 0 has no real solutions, thus the graph of this function has no 9-intercepts.

(4) Notice that 9 − 1 is the only factor in the denominator of the reduced function. Therefore, the only vertical asymptote will be 9 = 1. It is important to note that that cancelling the common factor 9 + 1 will not change the domain of the function, and it is still undefined at both 9 = 1 and 9 = −1. The graph will have a hole at 9 = −1.

Since d(−1 = ((�+(((�&�

(�(� = −1.5, the coordinates of the hole for this function are

(−1, −1.5).

(5) Observe that the degree of the numerator is one more than the degree of the denominator, therefore the graph of d will have an oblique asymptote. We will use synthetic division to find the quotient of (9� − 9 + 1 ÷ (9 − 1. 1 | 1 −1 1 1 0

1 0 1

Thus, d(9 = 9 + �

@(� , and the oblique asymptote is : = 9. We can solve d(9= 9 to

determine if the graph of dwill intersect the oblique asymptote.

@c&�@+(� = 9

9� + 1 = 9� − 9 9 = −1 We know that the function is undefined at 9 = −1; the graph will not intersect the oblique asymptote.

(6) Let us determine some additional points to helps us sketch the graph. Again, the table feature of the calculator is a useful tool for this step.

(7) d(−9 = (@c&�@+(� . The graph of d is not symmetric about the y-axis nor the origin.

(8) The graph follows.

x −3 −2 0.5 2 3

d(9 −3.25 −2.3 −1.5 3 3.5

40

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

y = x

1

1)(

2

3

−+=

x

xxf

Hole at (-1, -1.5)

x = 1

c. (1) d9 = @u&�@+ . The domain is the set of all real numbers except 0, or (�∞,0∪(0,∞.

(2) Function dis in lowest terms.

(3) Since x cannot equal 0, there is no :-intercept. On the other hand, since 90 � 2 = 0 has no real solutions, the graph of this function has no 9-intercepts.

(4) Solving 9� = 0, tells us that there will be a vertical asymptote at 9 =0.

(5) Notice that the degree of the numerator is two more than the degree of the denominator.

Dividing 90 + 2 by 9� results in d(9 = 9� + �@+, so, we have a non-linear quotient.

This tells us that the asymptote of this function will be : = 9�. We now solve d(9= 9� to determine if the graph of dwill intersect the oblique asymptote.

@u&�@+ = 9�

90+ 2 = 90

2 ≠ 0

There is no solution; the graph will not intersect the oblique asymptote.

(6) Determine some additional points to help sketch the graph.

(7) d(−9 = ((@u&�((@+ =

@u&�@+ . The graph of d is symmetric about the y-axis.

(8) The graph follows.

-5 -4 -3 -2 -1 1 2 3

-4

-2

2

4

6

8

10

12

x

y

2

4 2)(

x

xxf

+=

x = 0

y = x2

x −2 −1 −0.5 0.5 1 2

d(9 4.5 3 8.25 8.25 3 4.5

41

Finding a Rational Function

We have learned how to graph any rational function. Let us suppose we now want to reverse the process. That is, given the graph of a rational function, construct the equation.

For example, let us analyze the next graph to construct its equation.

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

(1, 1)

(0, 2/3)

y = 1

What do we know about this rational function? From the graph, we can see that it has vertical asymptotes at 9 = �2 and 9 = 3, which tells us that the denominator of the function will contain factors 9 � 2 and (9 − 3. The graph also shows 9-intercepts at 9 = −1 and 9 = 4. Recall that any 9-intercepts (zeros) of a rational function are found by solving the equation p(x) = 0, where p(x) is the numerator. Therefore, we know that the numerator of our rational function must contain the factors (9 + 1 and (9 − 4. Since the graph crosses the 9-axis, this tells us that these are zeros of odd multiplicity.

We can observe a horizontal asymptote at : = 1, which would represent the ratio of the leading coefficients. This implies that numerator and the denominator of the rational function have the same degree.

So far, we have the following possibility:

d(9 = >(@(0(@&�(@(�(@&�

We can see that the graph intersects the horizontal asymptote at (1, 1), thus, we can use this point to solve for a. Substituting 1 for d(9 = : and 1 for 9, we have

1 = >(�(0(�&�(�(�(�&�

1 = >((�(�((�(�

1 = a

So, our rational function can be d(9 = (@(0(@&�(@(�(@&� .

The given graph also shows a vertical intercept at (0, 2/3).

42

d0 = .(0.&�.(�.&� =

(0�(�� =

�� . It checks.

Thus, our rational function is d9 = @(0@&�@(�@&� or d9 =

92(�@(092(@(, . To check if our

answer is correct, we will graph this function with the calculator (standard window). See the graph that follows. Notice that it resembles that of the given graph.

Example 4

Construct an equation that corresponds to the given graph.

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-36-32-28-24-20-16-12

-8-4

48

12162024283236

x

y

(0, -16/3)

Hole at (-1, -9)

Solution

The graph of the rational function has a vertical asymptote at 9 = 3, thus the denominator of the rational function will contain the factor 9 � 3. The graph shows an 9-intercept at 9 = 2. Therefore, we know that the numerator of the function must contain the factor 9 � 2. Since the graph touches the 9-axis and turns around at 2, this is a zero of even multiplicity.

We can also observe a linear oblique asymptote. This implies that the degree of the numerator is one more than the degree of the denominator. Using any two points from the linear asymptote, say (1, 0) and (2, 4), we determine that the equation of the oblique asymptote is : = 49 � 4.

There is a hole at (�1, �9). This tells us that (x + 1) is a common factor of both numerator and denominator. So far, we have the following possibility:

d9 = >9�22@&�@(�@&�

We can use any point on the graph to solve for a. We will select (4, 16). Substituting 4 for 9 and 16 for d9 = :, we have

43

16 = >4�22(0&�(0(�(0&�

16 = >(0(�(�(�

16 = 4a

4 = a

Our rational function can be d(9 = 0(9−22(@&�(@(�(@&� .

The given graph also shows a vertical intercept at (0, −16/3).

d(0 = 0(0−22(.&�(.(�(.&� =

(0(0(�((� = − 16

3 . It checks.

Thus, our rational function is d(9 = 0(9−22(@&�(@(�(@&� or d(9 =

493(��92&�,92(�@(� .

The function graphed with the calculator confirms the equation.

[−3, 6, 1] by [−40, 40, 4]

The following screen includes the oblique asymptote.

[−3, 6, 1] by [−40, 40, 4]

In-Class Practice Section (D) 1. Which of the following functions has an oblique asymptote?

a. d(9 = @+(�@+(� b. d(9 =

@(�@+(� c. d(9 = @

+(�@(�

44

2. Sketch the graph of each rational function. Label any asymptote(s) and hole(s).

a. d9 = @+(�@(� b. d9 =

@c&-@+(0

3. Construct an equation that corresponds to the given graph.

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-10

-8

-6

-4

-2

2

4

6

8

10

12

x

y

(5, 5)

(0, -10/3)

x = -1 x = 3

Note: Refer to Section 7.2 for additional In-Class Practice exercises.

Exercises Section (D) 1. Which of the following functions has an oblique asymptote?

a. 2

2)(

2 +−=

x

xxf b.

2

2)(

2

−+=

x

xxf c.

65

365)(

2

2

−−+−=

xx

xxxf

In exercises 2-16, sketch the graph of each rational function. Show and label any asymptotes.

2. 1

54)(

2

−+−=

x

xxxf 3.

4

95)(

2

+−+=

x

xxxf 4.

6

18)(

2

+−+=

x

xxxf

5. 9

27)(

2

−+−=

x

xxxp 6.

5

8)(

2

+−=

x

xxh 7.

2

32)(

2

+−−=

x

xxf

8. 3

23)(

2

+−=

x

xxf 9.

2

14)(

2

−+=

x

xxf 10.

4

53)(

2

−+−=

x

xxxg

11. 2

27)(

2

+−=

x

xxxf 12.

5

32)(

2

−−−=

x

xxxf 13.

4

974)(

2

+++−=

x

xxxg

14. v9 = @u(@c&,@+ 15. d9 =

@u&0@c&�@+ 16.

1

24)(

2

24

−++=

x

xxxf

In exercises 17-22, using the asymptotes and given points as guides, construct the equation that corresponds to each graph.

45

17. 18.

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

y

(5, 4)

(-3, 0) (2, 0)

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

-20-18-16-14-12-10-8-6-4-2

2468

101214

x

y

(-3, -9)

(-1, 0)

(6, 0)

19. 20.

-14 -12 -10 -8 -6 -4 -2 2 4 6 8

-18

-15

-12

-9

-6

-3

3

6

9

12

x

y

(-5, -6.75)

(1, 0)

(4, 0)

-8 -6 -4 -2 2 4 6 8 10

-18

-15

-12

-9

-6

-3

3

6

9

12

x

y

(2, 9)

(-1, 0)

(-6, 0)

21. 22.

-8 -6 -4 -2 2 4 6 8 10

-18

-15

-12

-9

-6

-3

3

6

9

12

x

y

(4, 0)(0, 0)

(-6, 0) (6, -0.9)

-15 -12 -9 -6 -3 3 6 9

-60

-54-48

-42-36

-30-24-18

-12-6

6

121824

3036

x

y

(-1, 0)(-9, 12)

(6, 0)

(9, 0)x=1

In exercises 23-25, sketch the graph of each function. Label any asymptote(s) and hole(s).

23. 4

8)(

2

3

−+=

x

xxf 24.

12

1234)(

2

23

−+−−+=

xx

xxxxf 25.

9

1553)(

2

23

−−−−−=

x

xxxxf

Answer Key Section (D) Exercises

1. b

3. 5.

-14-12-10-8 -6 -4 -2 2 4 6 8 101214

-8

-6

-4

-2

2

4

6

x

f(x)

x=-4

y =x+1

-2 2 4 6 8 10 12 14 16

-16

-8

8

16

24

32

40

x

p(x)x=9

y =x+2

46

7. 9.

-16 -12 -8 -4 4 8 12 16

-30

-20

-10

10

20

30

x

f(x)

x=-2

y =-2x+4

-12 -9 -6 -3 3 6 9 12

-20

-10

10

20

30

40

50

x

f(x)

x=2

y=4x+8

11. 13.

-16 -12 -8 -4 4 8 12 16 20

-150

-120

-90

-60

-30

30

60

90

120

x

f(x)

x=-2

y =7x-16

-12 -8 -4 4 8 12

-120

-90

-60

-30

30

60

90

120

x

g(x)

y =-4x+23

x=-4

15.

-8 -6 -4 -2 2 4 6 8

-8

-4

4

8

12

16

x

f(x)

x=0

y =x2+4x

17. 43

6)(

2

2

−−−+=

xx

xxxf 19.

214

8102)(

2

2

−++−=

xx

xxxf 21.

64402

242)(

23

23

−−−−+=

xxx

xxxxf

23. 25.

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-18

-15

-12

-9

-6

-3

3

6

9

12

x

y

Hole at (-2, -3)

x=2

y = x

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-18-15-12-9-6-3

36

912

1518

x

y

Hole at (-3, 7/3)

x=3

y = -x-3

Note: Refer to Section 7.2 for applications of rational functions and additional exercises.

47

Section E: Nonlinear Systems of Equations

Solving by Graphing

From Chapter 2, we have learned how to solve linear systems in two variables and applied them to model real-world scenarios. We will now consider nonlinear systems of equations in two variables, that is, systems of equations in which there is at least one nonlinear equation. We can apply the same techniques we used for solving linear systems.

Solving nonlinear systems of equations graphically helps us visualize the graph of each of the equations in the system, so that we can explain the number of corresponding solutions. Just as with linear systems, we only need to draw the graphs of the two equations and see if and where they intersect. The solution to the system will be any point(s) that will satisfy both equations.

Example 1 Solve the system by graphing.

: = 29� − 9 : − 49 = −3

Solution

The graph of the system is shown below, and we see that there are two intersection points, which are (−1, −7) and (3, 9).

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-10-8-6-4-2

2468

101214

x

y y - 4x = -3y = 2x2 - 9

(-1, -7)

(3, 9)

To verify the solutions, we check that each ordered pair satisfies each equation of the system.

(−1,7: : = 29� − 9 −7 = 2(−1)� − 9 True : − 49 = −3 −7 − 4(−1) = −3 True

(3, 9): : = 29� − 9 9 = 2(3)� − 9 True : − 49 = −3 9 − 4(3) = −3 True

Thus, there are two real number solutions to the system: (−1, −7) and (3, 9).

A nonlinear system may have no solution, or 1, 2, 3, or even more solutions. Let us look at some examples of different possibilities.

48

No Real Solution One Real Solution

x

y

x

y

Two Real Solutions Four Real Solutions

x

y

x

y

Solving by graphing is not always very precise, especially when a point of intersection yields noninteger values, or when there are complex solutions. Using an algebraic method is usually more appropriate to help us find solutions of nonlinear systems of equations. Solving by Substitution

Recall that to solve a system by substitution we start by solving one of the equations for one of its variables in terms of the other. Then, we substitute the resulting expression into the other equation and we solve it. Once we solve for that one variable, we back-substitute to find the remaining coordinate. Basically, this method "converts" the system into one equation in one variable, which is more convenient to solve, and then we use back-substitution to find the value of the second variable.

The substitution method is very useful when solving a system that involves a linear equation and a nonlinear equation. Let us see an example in detail.

Example 2 Solve the system by substitution.

: � 39� = 5 : + 69 = 8

49

Solution

First, we solve one of the equations for one of its variables in terms of the other; we can choose any of the equations. Solving : � 69 = 8 for y, we get : = −69 + 8. We will replace the :of the nonlinear equation with the expression −69 + 8, then simplify and solve for x.

: + 39� = 5 Nonlinear equation

(−69 + 8 + 39� = 5 Let : = −69 + 8

39� − 69 + 3 = 0 Simplify and write in standard form.

3(9� − 29 + 1 = 0 Factor (GCF).

(9 − 1(9 − 1 = 0 Divide both sides by 3 and factor.

9 = 1 Apply zero-product property and solve for 9.

Now we can substitute 9 = 1 back into either of the original equations to find the value of :. From the linear equation, we know that : = −69 + 8, therefore, : = −6(1 + 8 = 2.

To verify the solution, we check that (1, 2) satisfies each equation of the system.

: + 39� = 5 2 + 3(1� = 5 True : + 69 = 8 2 + 6(1 = 8 True

The given system has only one (repeated) solution at (1, 2). Notice that a graphical method shows this solution.

-4 -3 -2 -1 1 2 3 4

-6

-4

-2

2

4

6

8

x

y

y + 6x = 8

y + 3x2 = 5 (1, 2)

! CAUTION

When solving nonlinear systems of equations it is important to check for any extraneous solutions. Always substitute the proposed solutions into each equation of the system and verify.

Example 3 Solve the system by substitution.

9� +:� = 2 : = −√9

Solution

Since : is already isolated on the second equation, we will substitute : = −√9 into the first equation, simplify, and solve for x.

50

9� �:� = 2 First nonlinear equation.

9� +(−√9� = 2 Let : = −√9.

9� + 9 − 2 = 0 Simplify and write in standard form.

(9 + 2(9 − 1 = 0 Factor.

9 = −2 or 9 = 1 Apply zero-product property and solve for 9.

We must immediately discard 9 = −2, since this value is not in the domain of : = −√9. Now we back-substitute 9 = 1 into either of the original equations to find the remaining coordinate. Using the second equation of the system, we have : = −√1 = −1.

Let us check that the ordered pair (1, −1) satisfies each equation of the system.

9� +:� = 2 (1� +(−1� = 2 True : = −√9 −1 = −√1 True The graph of the system shows the solution.

-3 -2 -1 1 2 3

-2

-1

1

2

x

y

(1, -1)

x y −=

x2 + y2 = 2

Example 4 Solve the system by substitution. Write your answer in fraction form as needed.

: − 3 = (9 − 5�

: = 9�

Solution

Substituting : = 9� into the first equation we get 9� − 3 = (9 − 5�. Now we simplify and solve for 9.

9� − 3 = 9� − 109 + 25 Square the binomial.

−3 = −109 + 25 Subtract 9� from both sides.

9 = �0� Solve for 9.

Back-substituting 9 = �0� into either of the original equations will give us the value of the

remaining coordinate. Using the second equation, we have : = )�0� *�= ��,

�� . The solution of the system is )�0� , ��,�� *. We proceed to check it.

51

: � 3 = 9 � 5� 7.84 � 3 " 2.8 � 5� True

: " 9� 7.84 " 2.8� True The graphical solution is shown below.

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-4

-2

2

4

6

8

10

12

14

16

18

x

y

y = x2 y - 3 = (x2 -5)2

25

1965

14,

[�10, 10, 1] by [�5, 20, 2]

Let us summarize the substitution method.

Solving a Nonlinear System in Two Variables by Substitution

(1) Solve one of the equations for one of the variables in terms of the other.

(2) Substitute the expression from step (1) into the other equation to get an equation in just one variable. Solve for the variable.

(3) Back-substitute the value found in step (2) into either of the original equations to find The value of the remaining variable.

(4) Check the solution(s) to the system in both original equations.

Solving by Elimination

The substitution method is convenient when one of the equations of the system has a variable that is already isolated, or at least one of the variables has a coefficient of 1 or �1, thus making it easier to isolate the variable. When none of these situations are given, it is more convenient to use the algebraic method of elimination.

Recall from Chapter 2 that we solve a system of equations by adding the two given equations, making sure that one of the variables drops out, leaving us with one equation in one unknown. Once we solve for that one variable, we back-substitute this value into one of the original equations to find the value of the remaining variable. Just as in the substitution method, it does not matter which variable you choose to eliminate first. The elimination method usually works very well when we have a nonlinear system where both variables are squared. Let us see an example. Example 5 Solve the system by elimination.

9�

9 � :�

4 " 1

9� �:� " 9

52

Solution

We can multiply both sides of the first equation by 36 to clear the fractions.

36)@+� �B+

0 * = 1(36) 49� �9:� " 36

So, we have the system 49� �9:� " 36

9� �:� " 9

Since neither variable will be eliminated when we add both equations, we can multiply the second equation by 9 to eliminate the :� terms. (It does not matter which variable we eliminate first; we could multiply the second equation by �4 instead to eliminate the 9� terms.)

9(9� �:� " 9) 99� � 9:� " 81

Make sure to multiply all the terms throughout the equation! Now we add the equations and solve for 9.

49� �9:� " 36

99� � 9:� " 81 139� �0:� = 117 9� = 9

9 " N3

! CAUTION

Before adding the equations, always make sure that the similar terms on both equations are vertically aligned. Finally, we back-substitute 9= �3 and 9 = 3 into one of the equations to find the :-value. Remember that you can choose any of the equations.

�3� �:� " 9 9 �:� " 9 : " 0

3� �:� " 9 9 �:� " 9 : " 0

The solutions are (�3, 0) and (3, 0). You can check that the solutions satisfy both equations of the system.

See the graph with the corresponding solutions.

x

y

(-3, 0)

x2 + y2 = 9

149

22=− yx

(3, 0)

[�15.16, 15.16, 1] by [�10, 10, 1]

Note: The graph of the first equation from Example 5 is called a hyperbola.

53

Following is summary of the elimination method.

Solving a Nonlinear System in Two Variables by Elimination

(1) Eliminate one of the variables by using the concept of opposites. If necessary, multiply one or both equations by a nonzero constant so that the coefficients of the terms of one of the variables are opposites of each other.

(2) Add the two equations, making sure that one of the variables drops out, leaving us with one equation in one unknown.

(3) Solve the resulting equation for the variable.

(4) Back-substitute the value found in (3) into one of the original equations to find the value of the remaining variable.

(5) Check the solution(s) to the system in both original equations.

Example 6 Solve the system by elimination. Round the coordinates of any solution(s) to 2 decimal places.

9� �:� = 9

9� � 4:� = 16

Solution

Multiplying both sides of the first equation by −1, we get−9� −:� = −9. We will add this equation to the second equation to eliminate the 9� terms.

−9� −:� = −9

9� + 4:� = 16 09� +3:� = 7

:� = 5�

: ≈ ±1.53

Now we back-substitute y = −1.53 and y = 1.53 into one of the equations to find the 9-value. We will select the first equation of the given system.

9� +(−1.53� = 9 9� +(1.53� = 9

9� = 6.6591 9� = 6.6591

9 = ±2.58 9 = ±2.58

The solutions are (−2.58, −1.53), (−2.58, 1.53), (2.58, −1.53), and (2.58, 1.53). We leave the check to you. The next graph displays the corresponding solutions.

54

x

y

(2.58, 1.53)

x2 + y2 = 9

x2 + 4y2 = 16(-2.58, 1.53)

(2.58, -1.53) (-2.58, -1.53)

Note: The graph of the second equation from Example 6 is called an ellipse.

Example 7 Jane and Diego own a company that develops applications for tablets. They focus on applications that do not require a large production investment. The cost, R, of their business is given by the function R9 = 859 + 40,000dollars, where 9 is the number of tablet applications sold. The revenue for sales of the tablet applications is modeled by w9 " 9� � 1759. Little Facts: Making a tablet application may take anywhere between two weeks to several months for the most complex ones, and there are many steps involved: design, coding, target audience, testing, infrastructure, platform, validation, management, distribution, etc. The production costs can vary between the low thousands to more than $100,000 for high-end applications. Sources: www.padgadget.com; www.tabletpublisherpro.com

a. How many tablet applications must the company sell to break even? b. Find the revenue at the break-even point. c. Find the cost at the break-even point. d. Graph the cost and revenue functions and show the break-even point.

Solution

a. The break-even point is where the cost and the revenue are equal, hence we only need to solve the system formed by the two given equations and determine wherew(9 = R(9.

9� + 1759 = 859 + 40000 Substitute for w(9 and R(9. 9� + 909 −40000 = 0 Simplify and write in standard form.

(9 − 160(9 +250 = 0 Factor.

9 = 160 or 9 = −250 Apply zero-product property and solve for 9.

Since the break-point implies nonnegative numbers, we can discard 9 = −250, and the solution is 9 = 160. Thus, the company must sell 160 tablet applications to break even.

b. w(160 = (160� + 175(160 = $53,600.

c. R160 = 85(160 + 40,000 " $53,600.

d. The graph follows.

55

-200 200 400 600 800 1000

10000

20000

30000

40000

50000

60000

70000

x

$

C(x) = 85x +40,000

R(x) = x2 + 175x

Break-even point(160, 53600)

Example 8

Solve the given systems using any of the methods discussed. Round the coordinates of any solution(s) to 4 decimal places.

a. : = √9 b. : = log9 � 3 � 1

: = � log9 � 3 � 2 : = � log9 � 6 � 2

Note: Refer to Chapter 6 to review logarithms.

Solution

a. Solving graphically, we determine that there is one point of intersection at (1.7509, 1.3232).

[-5, 5, 1] by [-5, 5, 1]

Checking the solution, we have 1.3232≈√1.7509 True and 1.3232≈ � log1.7509 � 3 � 2 True

b. Solving by substitution will result in a logarithmic equation that we can solve to find any solution(s) of the system.

log9 � 3 � 1 " � log9 � 6 � 2

log9 � 3 � log9 � 6 " 1 Collect the logarithmic expressions on one side.

log9 � 39 � 6 " 1 Product property.

9 � 39 � 6 " 10� Convert to exponential form.

9� � 99 � 18 = 10 Distributive property; simplify left-hand side.

9 � 19 � 8 = 0 Set equal to 0 and factor.

9 = �1 or 9 = �8 Apply zero-product property and solve for 9.

We must discard 9 = �8 because it is not in the domain. Finally, we back-substitute

56

9 = �1 into one of the equations of the system to find the :-value. : = log[�1 � 3< � 1 = log2 �1≈1.3010. The exact solution is (�1, log (2) + 1) or approximately (�1, 1.3010). Using the graphing calculator, we can verify the solution by the intersection method.

[-5, 5, 1] by [-5, 5, 1]

Nonlinear Systems with Complex Solutions

Consider the graphs of the equations that form the following system.

: � 0.259� " 5

: � 0.259� " �3

-14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14

-35-30-25-20-15-10

-5

5101520253035

x

y

y - 0.25x2 = 5

y + 0.25x2 = -3

Observe that the graphs do not intersect. This is an example of a system that has no real-number solutions. Let us analyze the algebraic solution of this system.

We see that the coefficients of 9� are opposites, thus we can add the given equations to eliminate the 9� terms.

: � 0.259� " 5

: � 0.259� " �3

2: �09� " 2

2: " 2

: " 1

Now we back-substitute : = 1 into any of the equations to find the 9-value; we will use the second equation.

1 � 0.259� " �3 0.259� " �4 9� " �16 9 " N√�16 " 4�

Notice that there are two complex solutions, (�4�, 1) and (4�, 1), which will not appear as intersecting points on the graph of this system; there are no real-number solutions.

57

Checking:

: � 0.259� = 5 1 − 0.25(−4�� = 5 1 − 0.25(−16 = 5 True

: + 0.259� = −3 1 + 0.25(−4�� = −3 1 + 0.25(−16 = −3 True

: − 0.259� = 5 1 − 0.25(4�� = 5 1 − 0.25(−16 = 5 True

: + 0.259� = −3 1 + 0.25(4�� = −3 1 + 0.25(−16 = −3 True

Example 9 Solve the system algebraically.

: +39� = 2

: + 29 = 3

Solution

Solving the second equation for : results in : = −29 + 3. We will substitute : = −29 + 3 into the first equation, simplify, and solve for 9.

(−29 + 3 + 39� = 2 Let : = −29 + 3

39� − 29 + 1 = 0 Simplify and write in standard form.

Since 39� − 29 + 1 = 0 is not factorable, we will use the quadratic formula.

x = (?±√?+(0>|

�> with a = 3, b = −2, and c = 1.

x =(((�±�((�+(0(�(��(� = �±√(-

, = �±�'√�, = �±'√�

�

x = �&'√�� and x = �('√�

� or x = �� +

√�� i and x = �� −

√�� i .

Back substituting the 9-values into : = −29 + 3, we get

: = −2)�� +√�� �* + 3 = 5� −

�√�� � and : = −2)�� −

√�� �* + 3 = 5� +

�√�� �.

The two complex solutions are )�� +√�� �, 5

� ��√�� �* and )�� �

√�� �, 5

� ��√�� �*.

Observe that the graphs of the equations do not intersect.

-2 -1 1 2 3

-3

-2

-1

1

2

3

x

y

y = -3x2 + 2 y = -2x + 3

58

In-Class Practice Section (E) 1. Given the graphs of the equations of a nonlinear system, find and verify the solutions.

-5 -4 -3 -2 -1 1 2 3 4 5

-3

-2

-1

1

2

x

y

y = -2x2 + 4x -2

(x - 1)2 + (y + 1)2 = 2

2. Solve each system graphically and verify any solution(s).

a. 9� �:� = 2 b. : = −2@ + 4 : = 9� : = log(9 + 10 + 2

3. Solve the system by elimination or substitution. Show the graphical solution(s).

a. 9� +:� = 9 b. 9� +:� = 4 c. 9� +:� = 16 : = −9 + 3 29� +:� = 8 9� +:� = 25

4. Small planes regularly fly over an open area that contains a popular tourist attraction whose shape is that of a parabola. The attraction's structure can be modeled by the equation : = −0.006259� + 49. Suppose a small plane departs on a northwest direction from a nearby non-commercial airport. The plane's flight path is denoted by the equation : = −29 + 1240. Determine if there is any danger that the plane will crash into the popular tourist attraction. Solve algebraically and include the corresponding graph. Little Facts: The Federal Aviation Administration (FAA) is the national aviation authority of the U.S.A. The FAA establishes that except when necessary for takeoff or landing, no person may operate an aircraft over any congested area of a city, town, or settlement, or over any open air assembly of persons, below an altitude of 1,000 feet above the highest obstacle within a horizontal radius of 2,000 feet of the aircraft. In areas that are not considered congested, the aircraft cannot go below an altitude of 500 feet above the surface except over open water or sparsely populated areas; the aircraft may not be operated closer than 500 feet to any person, vessel, vehicle, or structure. Source: www.faa.gov

Exercises Section (E) In exercises 1-8, solve the system by graphing. Round answers to 2 decimal places where needed.

1. 42

162 2

−=+−=

xy

xy 2.

093

1032

=+−−+=

xy

xxy 3.

1

2522

−−==+

xy

yx 4.

64

02

38

−=

=++

xy

xy

5. 102

2

−+=

−=

xy

xy 6.

9� + :� = 9: = |9| + 0.5 7.

52

1

16

2

22

+−=

=+

xy

yx 8.

4)8log(

43

++=−=xy

y x

59

In exercises 9-16, solve each system using substitution. Round your answer(s) to 2 decimal places in ordered pair form where needed.

9. 53

32 2

=+−=−

xy

xy 10.

1

)5(22

2

−=+=−

xy

xy 11.

6

)1(42

2

−=−+−=−

xy

xy 12.

)2log(1

)2

1log(

xy

xy

−=

−=

13. )3log(1

)log(

−−==

xy

xy 14.

6

9322

22

=+=+

yx

yx 15.

94

8222

22

=+=+

yx

yx 16.

)4ln(

)2ln(2 −=−=

xy

xy

In exercises 17-24, solve each system using elimination. Round your answer(s) to 2 decimal places in ordered pair form where needed.

17. 22

18222

22

=+=+

yx

yx 18.

213

522

22

=+−=+−

yx

yx 19.

8

1022

22

=+−=+

yx

yx 20.

24

6322

22

=+=+

yx

yx

21. 3032

122

2

=+−=−

yx

yx 22.

2

12 2

=−−=+−

yx

yx 23.

7

174 2

=+−=−

yx

yx 24.

53

02 2

=−−=+

yx

yx

25. Carlos is an art teacher who incorporates math into his class. He asks his students to draw a circle given by the equation .49416 22 =+ yx He then asks them to place a pencil on the circle that denotes the linear equation .32 += xy Find the points where the student's pencil intersects the circle. Write answers in ordered pair form, rounding to 2 decimal places.

26. A group of students are working on a homework problem in which a laser beam crosses through the path of a particle orbiting an object centered at the origin. Suppose the laser beam follows a path given by 43 −−= xy , and the path of the particle orbiting the object is

given by .369 22 =+ yx Find the points where the laser beam intersects the path of the orbiting particle. Write your answers in ordered pair form, rounding to 2 decimal places. Little Facts: The word Laser is an acronym for Light Amplification Stimulated Emission Radiation. The first successful optical or light laser was considered to be the “ruby laser” invented by Theodore Maiman in 1960. Some speculate though, that is was Gordon Gould who first used the word laser and that he was the one who made the first light laser. Sources: www.inventors.about.com; www.press.uchicago.edu

27. Mark, Randy, and Steven are college students on Spring break, relaxing on the beach playing Frisbie. Steven decides to sit out on a game of Frisbie played by Mark and Randy. Assume that Steven is sitting at the point of origin. Mark and Randy throw their Frisbees at the same time, with Randy’s Frisbie following a path given by 52 2 −= xy and Mark’s Frisbie given by .34 −= xy Assuming that the Frisbees are thrown back given the same path, find the points where the Frisbees intersect. Write your answers in ordered pair form, rounding to 2 decimal places. Little Facts: The Bridgeport Connecticut Frisbie Pie Company sold pies to New England colleges in the late 1870’s. Throwing and catching the empty pie tins became popular to Yale college students for past time recreation. It was Russel Frisbie who put the family name on the bottom of the pie tins that made the homemade pies and their tins popular. Sources: www.catchthespirit.co.uk; www.wfdf.org; www.ideafinder.com

60

Answer Key Section (E) Exercises 1. (�3,2); (2, −8) 3. (3, −4); (−4,3

-5 -4 -3 -2 -1 1 2 3 4 5

-18-16-14-12-10-8-6-4-2

246

x

y

y = 2x2-16y = -2x- 4

(-3,2)

(2, -8)

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

y = x2+3x-10

y= 3x - 9

(-4, 3)

(3, -4)

5. (11, 3) 7. (−1.65, 3.65); (1.65, 3.65)

-2 2 4 6 8 10 12 14 16

-4

-2

2

4

6

8

10

x

y

2−= xy

(1, -6)

(11, 3)

102 −−−−++++==== xy

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9

-5-4-3-2-1

12345

x

y

(-1.65, 3.65)

x2 + y2 = 16

5

2

1 2 +−= xy

(1.65, 3.65)

9. (−2.89, 13.66; (1.39, 0.84) 11. (−2.68, 1.18; (1.68, �3.18) 13. (5, 0.7 15. (0.71, 2.64); (−0.71, 2.64; (0.71, −2.64; (−0.71, �2.64

17. (5.1, 2�); (−5.1, 2�; (5.1, −2�; (−5.1, �2� 19. (1, 3; (−1, 3; (1, −3; (−1, �3

21. (3, 2; (3, �2 23. (2, 5; (16, �9 25. (�1.73, �0.46; (0.23, 3.47

27. (−0.41, �4.66; (2.41, 6.66

61

Section F: Other Types of Equations

Equations Reducible to Quadratic Form

In Chapter 4 we learned how to solve quadratic equations by different algebraic methods, as well as graphically and numerically. In this section, we will examine equations that, although not exactly quadratic, can be made to look like quadratic, or be reduced to quadratic form. We call these equations quadratic in form or reducible to quadratic. These equations can be solved by applying the techniques we have used for solving quadratic equations.

For example, suppose we have the equation 90 � 69� � 8 " 0. This equation is not quadratic in 9, but notice it would be quadratic in 9�. We can write the equation as 9�� � 69�� � 8 = 0.

Suppose we let } = 9�, then we have }� = 90. So, the given equation becomes the following quadratic equation in }, which we can now solve by factoring.

}� � 6} � 8 " 0 Substitute: } = 9� and }� = 90.

} � 4} � 2 = 0 Factor.

} − 4 = 0 or } − 2 = 0 Apply zero-product property.

} = 4 or } = 2 Solve for }.

Observe that these are not the solutions of the original equation in terms of 9 (our goal is to find the values of 9, not values of }). Therefore, we now use reverse substitution to solve the equation in terms of the original variable, 9. Since } = 9�, we have

9� = 4 9� = 2

9 = ±√4 9 = ±√2

The solutions are 9 = −2, 2, −√2, and √2. Let us check the solutions in 90 − 69� + 8 = 0. (−20 − 6(−2� + 8 = 0 16 − 24 + 8 = 0 True (20 − 6(2� + 8 = 0 16 − 24 + 8 = 0 True(−√20 − 6(−√2� + 8 = 0 4 − 12 + 8 = 0 True(√20 − 6(√2� + 8 = 0 4 − 12 + 8 = 0 True

We can check the answers graphically. Notice the zeros at the solution values.

-2 -1 1 2

-2

2

4

6

8

x

y

)0,2()0,2(−−−− )0,2(− )0,2(

Equations Quadratic in Form

An equation is said to be quadratic in form if it is reducible to a quadratic equation that can be written as

�}� + �} + = 0, where a, b, and c are real numbers, a ≠ 0, and } is an algebraic expression.

62

Example 1 Solve using substitution.

a. 9 � 4� �⁄ � 9 � 4� �⁄ " 2 b. 9 � 3√9 � 4 " 0

Solution

a. Let } = 9 � 4� �⁄ . The expression 9 � 4� �⁄ is equivalent to ;9 � 4� �⁄ <�. Now we write the given equation as a quadratic equation in standard form in terms of }, and solve for }.

}� � } � 2 " 0 Substitute }� for 9 � 4� �⁄ and } for 9 � 4� �⁄ .

} � 2} � 1 " 0 Factor.

} � 2 " 0 or } � 1 " 0 Apply zero-product property.

} " �2 or } " 1 Solve for }.

! CAUTION

Remember to reverse substitute to solve the equation in terms of the original variable, 9.

Now we will replace } with 9 � 4� �⁄ and solve for 9.

} " �2 } " 1

9 � 4� �⁄ " �2 9 � 4� �⁄ " 1

;9 � 4� �⁄ <� " �2� ;9 � 4� �⁄ <� " 1�

9 � 4 " �8 9 � 4 " 1

9 " �12 9 " �3

Let us check the solutions in the original equation, 9 � 4� �⁄ � 9 � 4� �⁄ " 2. �12 � 4� �⁄ � �12 � 4� �⁄ " 2 �8� �⁄ � �8� �⁄ = 2 4 � 2 " 2 True �3 � 4� �⁄ � �3 � 4� �⁄ " 2 1� �⁄ � 1� �⁄ " 2 1 � 1 " 2 True

Since both check, the solutions are 9 " �12 and 9 " �3. We can see that the graph of the equation displays the zeros at these values.

[�15, 4, 3] by [�4, 4, 2] [�15, 4, 3] by [�4, 4, 2]

b. If we let } = √9 in 9 � 3√9 � 4 " 0, then }� " 9. We now write the equation in terms of }.

63

}� � 3} − 4 = 0 Substitute.

(} + 4(} − 1 = 0 Factor.

} = −4 or } = 1 Apply zero-product property and solve for }.

Replacing } with √9 , we get √9 = −4 or √9 = 1. There is no real solution for √9 = −4 , and we must discard it. On the other hand, if √9 = 1, then 9 = 1, which is the only real solution to the original equation. Notice that the graph shows only one zero, at 9 = 1.

-1 1 2 3

-5-4-3-2-1

12345

x

y

(1, 0)

Example 2 Solve using substitution.

a. 9(� − 39(� = 10 b. 90 − 129� − 64 = 0

Solution

a. Subtracting10 from both sides, we have 9(� − 39(� − 10 = 0. Letting } = 9(�, we know that }� = (9(�� = 9(�. Now we write the given equation as a quadratic equation in terms of }, and solve for }.

}� − 3} − 10 = 0 Substitute.

(} − 5(} + 2 = 0 Factor.

} = 5 or } = −2 Apply zero-product property and solve for }.

We reverse substitute to solve the equation in terms of the original variable, 9.

9(� = 5 9(� = −2

9 = �� 9 = − �

� 9(� is equivalent to the reciprocal of 9.

You can verify that both solutions will check. Observe both real solutions on the graph.

-3 -2 -1 1 2 3

-12-10-8-6-4-2

2468

10121416

x

y

(1/5, 0)(-1/2, 0)

64

b. If we let } = 9�, then }� = 90. The equation 90 � 129� − 64 = 0 becomes the following quadratic equation in }, which we can now solve by factoring.

}� − 12} − 64 = 0 Substitute.

(} − 16(} + 4 = 0 Factor.

} = 16 or } = −4 Apply zero-product property and solve for }.

Applying reverse substitution, we have

9� = 16 9� = −4

9 = ±√16 = ±4 9 = ±√−4 = ±2� Hence, we have the following solutions to the equation: 9 = −4, 9 = 4, 9 = −2�, and 9 = 2�. We can see that only the real solutions will be displayed on the graph.

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-80

-60

-40

-20

20

40

x

y

(4, 0)(-4, 0)

Rational Equations

A rational equation is an equation that has one or more rational expressions. Some examples are:

��@ + 3 =

��@ ,

�@&� = 2 +

�@&�,

@+

@(0 = �,

@(0

Let us review how to solve rational equations.

Solving Rational Equations

(1) Clear the equation of fractions by multiplying both sides by the LCD.

(2) Solve the resulting equation.

(3) Check the solution(s) in the original equation. Exclude any values for which any rational expression is undefined (that is, any value that would yield a zero denominator).

Example 3 Solve the rational equation.

��@ + 3 =

��@

65

Solution

The LCD is 109. The first step is to multiply both sides of the equation by 109 to clear the fractions. Then, we can solve the resulting equation.

(109)) ��@ � 3* = ) �

�@*(109) Multiply each side of the equation by the LCD.

(109)) 259* � (109)(3) " ) ��@*(109) Apply the distributive property.

4 + 309 "5 Simplify.

309 " 1 Subtract 4.

9 " ��. Solve for 9.

The rational expressions in the given equation are undefined only if 9 = 0. Let us check the solution in the original equation. ��@ + 3 "

��@

��) Aco*

+ 3 " ��) Aco*

12 + 3 = 15 True

! CAUTION

Make sure you multiply all terms of the equation by the LCD when clearing the fractions.

Example 4

Solve each rational equation.

a. �

@&� " 2 + �@&� b.

@+@(0 "

�,@(0

Solution

a. We first multiply both sides of the equation by the LCD = 9 � 1 to clear the fractions.

(9 � 1) �

@&� " 2(9 � 1) � �@&�(9 � 1) Multiply all terms of the equation by the LCD.

3 " 29 + 7 Simplify and combine like terms.

�4 " 29 Subtract 7

�2 " 9 Solve for 9. Since 9 " �2 does not make any denominator zero, it will check.

�

@&� " 2 � �@&�

�(�&� " 2 � �

(�&� �3 = 2 �5 True

b. Multiplying both sides of the equation by the LCD = 9 � 4 we have

9 � 4 ) 929�4* " ) �,

@(0* 9 � 4 9� " 16

Therefore, 9 " N4. Since we must reject any value of 9 that yields a zero denominator, 9 " 4 must be discarded, and the solution for the given equation is9 " �4. Furthermore,

66

notice that 9 = 4 is not in the domain of @+@(0 or

�,@(0 .

You can check it algebraically.

Example 5

Solve the equation. (�

@+(0@ � ,@ "

�@(5@(0

Solution

The factors of the denominator 9� � 49 are 99 � 4. Considering all denominators, the LCD will be 99 � 4. Let us multiply both sides of the equation by the LCD to clear the fractions.

99 � 4)U �999�4V � )69* 99 � 4) " )29�79�4 * 99 � 4)

This will result in the following equation, which will lead us to a quadratic equation.

�9 � 69 � 4 " 29 � 79 The solution follows.

�9 � 69 � 24 " 29� � 79 Apply the distributive property.

� 69 � 15 " 29� �79 Combine like terms.

29� �9 � 15 " 0 Write the equation in standard form.

29 � 59 � 3 " 0 Factor.

9 " � 52 or 9 " 3 Apply zero-product property and solve for 9.

Since the original restrictions are 9 � 0 and 9 � 4, both solutions will check. To confirm the solutions, we can also graph and solve by the Intersection method, as shown next.

�1 " (�

@+(0@ � ,@ �2 " �@(5

@(0

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

(3, 1)

Y1

(-5/2, 24/13)

Y2

67

Example 6

Solve the equation for . ?|

|&j = � Solution

We are now solving a literal equation with a rational expression. We assume that � ! ≠ 0. Let us clear the fractions by multiplying both sides of the equation by the LCD " � !. � ! ) ?|

|&j* " � � !

This will result in the following: � " � � �!

Solving for , we have

� � � " �!

� � � " �!

" >j

?(>

Example 7 Lynn is buying food from a catering service for a neighborhood block party. She bought 20 pounds of food items that consisted of various salads at $4 per pound, and meats at $8 a pound. Use the following equation to determine how many pounds of each food item she bought, if 9 represents the pounds of salads.

�,@ "

,0�.(@

Solution

First, we multiply both sides of the rational equation by the LCD = 920 − 9 to clear the fractions.

(9(20 − 9 )�,@ * = ) ,0�.(@* (9(20 − 9

Next, we will solve the resulting equation:96(20 − 9 = 649.

1920 − 969 = 649

1920 = 1609

12 = 9

The rational expressions in the given equation are undefined only if 9 = 0 or 9 = 20. Since 9 = 12, it checks.

We know that 9 represents the pounds of salad, thus 20 − 9 will give us the pounds of meats. Therefore, Lynn bough 12 pounds of salads and 8 pounds of meats.

68

Background for Example 8: The rate of work is the number of tasks that can be completed in a given unit

of time. If a task can be completed in �units of time, the rate of work is equal to �� of the task per unit of

time.

Example 8 Kevin and Juan have a part-time job painting patio playhouses. Kevin can paint a playhouse in 5 hours. If Juan and Kevin work together, they can paint a playhouse in 2 hours. How long would it take Juan to paint a playhouse working alone?

Solution

Let 9 = Juan's time (in hours) to paint a playhouse working alone. Then we know that

�� = Kevin's rate of work, and

�@ = Juan's rate of work. Multiplying each person's rate by the time

it takes both to paint a playhouse working together will give us each painter's portion of the task. Kevin's portion of the task + Juan's portion of the task = One completed task

)��*(2) + )�@*(2) = 1

Solving the equation for 9, we have

59 )��*+)�@* 59 = 159

29+10 = 59

�.� = 9

Therefore, it would take Juan 3 �� hours (or equivalently, 3 hours and 20 minutes) to paint a

playhouse working alone.

In-Class Practice Section (F) 1. Solve using substitution to find any real solutions.

a. 90 − 79� − 18 = 0 b. 39� �⁄ − 9� �⁄ = 2 c. 29 + 9√9 = −9

d. 9(� + 59(� = 24 e. √39� − 2u = 9

2. Confirm the solutions to problems 1(a) and 1(b) graphically.

3. Solve each rational equation. State any restricted value(s).

a. ,

�@&� = 3 + 0�@&� b.

@+

@(� = ��

@(� c. �@(�

� + �@@&� = 9

4. Solve the equation for 9: �@ +

�B = 1. Assume 9, : ≠ 0

5. Sophie can develop a science project in 6 hours. Brent needs 8 hours to prepare the same project. How long will it take them to develop the science project if they work together? Round your answer to the nearest tenth. Answer in a complete sentence.

69

Exercises Section (F)

In exercises 1-18, solve using substitution to find any real solutions. Round to 2 decimal places as needed.

1. 128 24 −=− xx 2. 24 546 xx −=− 3. 0144 24 =++ xx

4. 05129 24 =−+ yy 5. 65 24 += yy 6. 5027 3/13/2 −=+ xx

7. 0168 3/13/2 =− xx 8. 132 4/12/1 −=− xx 9. 0561330 4/12/1 =−+ xx

10. 12712 =+ xx 11. 615 +−= xx 12. 8233 2 =− xx

13. 12 736 −− =− xx 14. 12 1428 −− = yy 15. 21)2(4)2( 3/13/2 =+−+ yy

16. 02)5()5( 4/12/1 =−−−− xx 17. 12)70()70( 4/12/1 =+++ xx 18. 3/16/1 127 xx −−=−

In exercises 19-26, solve each rational equation. State any restricted value(s). Write answer(s) in fraction form where needed.

19. 14

1

3 −=−+ xx

20. 4

9

3

210 =+

+xx

21. 4

7

1

3

2

3 =+

+xx

22. 5

13

2

2

4

32

=−

+− xx

23. 9

8

5

1

3

1 =+

−− xx

24. 5

6

5

6122 −

−=−

+−xxxx

25. 4

18

5

7

+−=

−+

xx

x 26.

1

2

2

8

−−=

+−

xx

x

27. Professor Nagamo has an English degree with a minor in Physics. He picks 2 students, Rich and Chase, to write a biography on his favorite physicist Michio Kaku, who is also of Japanese descent. The students are then to present their paper(s) along with a short speech on what they learned about Kaku and the subject of physics. Assume that time represents hours spent on researching, editing, typing the paper and printing it out. Rich takes 4 hours whereas Chase takes 6 hours to complete the paper. How long would it take if they worked together? Round your answer to 1 decimal place. Little Facts: Dr. Michio Kaku is a theoretical physicist that is popular amongst college students due to his knack of explaining complicated science in a way that makes it understandable and interesting. He has also made appearances on the Colbert Report, Discovery, CNN, BBC, and the Science Channel. As of 2012, Dr. Kaku, a co-founder of string field theory, continues his research on the unified field theory which Einstein had theorized but had not proved. Sources: www.mkaku.org; www.aps.org

28. Bill and Will are brothers who are both taking graduate level art classes. To help pay off their student loans, they decide to sell watercolor replicas of Andy Warhol’s art. To produce one piece of work, Bill takes 0.25 hours less than Will. Working together, it takes them 3 hours. Find the time it would take each of them if they worked individually. Round your answers to 2 decimal places. Little Facts: Andy was born on August 6th, 1928 in Pittsburg, Pennsylvania to Carpatho-Rusyn immigrant parents. His passion began at the age of nine when he was given his first camera. He took a free art class at Carnegie Institute while still in high school, eventually earning a Bachelor of Fine Arts in Pictorial design at Carnegie Institute of Technology, which is now known as Carnegie Mellon University. His success first began when his work debuted in Glamour magazine in 1949. Warhol is most notable for his silk screen depictions of

70

Marilyn Monroe and Campbell soup cans. Source: www.warhol.org 29. After attending a 20 year high school reunion, the group of former classmates decides to go to a local amusement park along with their children. The price for each adult ticket was $76, and the children tickets were $38 apiece. Let 9 represent the number of children tickets. Use the equation below to determine how many of each were bought if there were a total 60 tickets sold to the group.

��0.@ =

�0�.,.(@

In exercises 30-33, solve each equation for the given variable.

30. wzx

xy =−

; solve for x 31. 2)( =

+−pn

pnm; solve for n

32. yx

xy =+−++5

13

4

32; solve for x 33.

bababa

ab

+=

−+

−24

22; solve for b

Answer Key Section (F) Exercises

1. 45.2± ; 41.1± 3. No real solutions 5. 45.2± 7. 0; 8 9. 1.85 11. 0.36

13. 0.67; –3 15. 341; –29 17. 11 19. 2±=x ; 1,0−≠x 21. 7/3,2−=x ; 1,0−≠x

23. 6,4−=x ; 5,3−≠x 25. 31,2−=x ; 5,4−≠x 27. 2.4 hours

29. Children: 15 tickets; Adults: 45 tickets 31. m

pmpn

−−−=

2

2 33.

a

ab

+−=6

2

71

Section G: Solving Polynomial and Rational Inequalities

In Chapters 2 and 3 we worked with linear and absolute value inequalities, respectively. In Chapter 4 we learned how to solve quadratic inequalities, both graphically and algebraically. Now we will solve polynomial inequalities in general and rational inequalities.

Polynomial Inequalities

Polynomial Inequalities

A polynomial inequality is an inequality which can be written as d9 < 0, d9 > 0, d9 ≤ 0, or d9 ≥ 0, where d9 is a polynomial function.

Solving by Graphing

Graphs are useful to help us visualize the solutions of a polynomial inequality. Let us revisit how to solve a second-degree polynomial by graphing, for example, 9� � 9 � 6 < 0. Recall that the solutions of a quadratic equation of the form ax2 + bx + c = 0 occur at the zeros of the graph of d9 = ax2 + bx + c, where the graph may have one, two, or no 9-intercepts.

We can solve the inequality 9� � 9 � 6 < 0 by examining the graph of the associated function d9= 9� � 9 � 6and finding the intervals where the function is below the 9-axis. So, we will graph d9 = 9� � 9 � 6 and locate the zeros.

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-6

-4

-2

2

4

6

8

x

y

The zeros are (�2, 0) and (3, 0). We can see that the graph is below the axis when �2 < 9 < 3. Hence, the solution set of 9� � 9 � 6 < 0 contains all 9-values in the interval (�2, 3). See the blue portion on the next graph.

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-6

-4

-2

2

4

6

8

x

y

72

Example 1 Solve the following polynomial inequalities by graphing.

a. 9� � 29� − 9 − 2 ≤ 0 b. 490 − 89� − 39� ≥ −59 − 2 c. 290 + 9� + 9 + 2 < 0 Solution

a. We first graph the associated functiond(9= 9� + 29� − 9 − 2and find the zeros.

-4 -3 -2 -1 1 2 3 4

-8

-6

-4

-2

2

4

6

8

x

y

The zeros are (−2, 0), (−1, 0), and (1, 0). The graph of the polynomial is on or below the 9-axis for 9 ≤ −2, and when −1 ≤ 9 ≤ 1. See the graph below.

-4 -3 -2 -1 1 2 3 4

-8

-6

-4

-2

2

4

6

8

x

y

The solution in interval notation is given by (−∞,�2< ∪ [−1,1<. b. Rewriting the inequality as 490 − 89� − 39� + 59 + 2 ≥ 0 and graphing the associated function, we observe that there are three 9-intercepts: (−0.5, 0), (1, 0), and (2, 0).

-2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5

-4

-2

2

4

6

8

10

x

y

As shown next, the graph of the polynomial is on or above the 9-axis for 9 ≤ 1, and 9 ≥ 2.

73

-2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5

-4

-2

2

4

6

8

10

x

y

So, the solution for the inequality490 � 89� � 39� + 59 + 2 ≥ 0, or equivalently 490 − 89� − 39� ≥ −59 − 2, is (−∞,1< ∪ [2, ∞. c. The graph of d9 = 290 + 9� + 9 + 2 is shown below.

-2 -1 1 2

-4-2

2468

101214

x

y

Since the graph will never cross the 9-axis, it can never be less than 0. Thus, this inequality has no real solutions.

Example 2 Solve the following polynomial inequalities with the calculator. Round your answers to 2 decimal places as needed.

a. 49� + 59� − 89 > 0 b. 59� − 129� ≤ 29 +3

Solution

a. We will solve the inequality using the Zero feature of the graphing calculator. First, we graph d(9 = 49� + 59� − 89, then find the 9-intercepts, and determine the portion of the graph that lies above the 9-axis. Using the window [−5, 5, 1] by [−8, 12, 2], the zeros are displayed.

The intervals that contain the solutions of 49� + 59� − 89 > 0 are (−2.17, 0) and (0.92,∞).

74

b. We will solve 59� � 129� � 29 �3 by using the Intersect feature of the graphing calculator. Letting Y1 = 59� � 129� and Y2 = 29 �3, we will find where Y1 = Y2 and determine where Y1 � Y2. We can interpret Y1 � Y2 as the values of 9 where the graph of Y1 is below or equal to the graph of Y2.

Y2 = 29 �3 Y1 = 59� � 129�

Notice that 59� � 129� � 29 �3 when 9 � 2.64 (rounded to 2 decimal places), or the corresponding interval, (�∞,2.64].

Example 3 The number (in thousands) of plant-eating insects infesting an abandoned garden, before a group of neighbors joined forces to restore it, is modeled by�9 " �9 � 19 � 29 � 3� � 4, where 9 represents the number of weeks. When was the plant-eating insects' population greater than 3,500? Round your answer to the nearest whole number.

Solution

We will solve�9 � 3500 by using the Intersect feature of the graphing calculator. Let Y1 = �9 � 19 � 29 � 3� � 4and Y2 =3.5. Using the window [0, 5, 1] by [0, 5, 1], we will determine where Y1 � Y2.

Y1 Y2 Y1 Y2

Observe that that �9 � 3500 when 9 � 0.8974 or 9 < 3.3883. Rounded to the nearest whole number, we can say that the plant-eating insects' population was greater than 3,500 after the first week and before the third week.

Solving Algebraically

We can solve a polynomial inequality by finding the zeros algebraically, testing the intervals around the zeros in the inequality, and then verifying any 9-values that make the inequality true. Let us review (from Chapter 4) how to solve a quadratic inequality algebraically, for example, 9� � 9 � 1 � 7.

Moving all the terms to one side of the inequality and leaving 0 on the other we have

9� � 9 � 6 � 0.

75

Let us replace the inequality symbol with an equal sign and solve the related equation by factoring.

9� � 9 � 6 = 0

9 � 39 � 2 = 0

9 � 3 = 0 or 9 � 2 = 0

9 = �3 or 9 = 2Now we plot these solutions on a real number line, dividing the number line into three intervals. The real solutions we found, 9 = �3 and 9 = 2, are usually referred to as critical values or boundary points. I II III

x

y

0 -3 2 �∞,�3 �3, 2 (2, ∞

We use an open circle to indicate that our inequality, 9� � 9 � 6 � 0, is not satisfied at these points. (If we have an inequality with ≤ or ≥ we use the solid dot; if the inequality contains < or >, then we use an open circle.)

To solve our inequality, we need to confirm which of the numbers within each interval will satisfy 9� � 9 � 6 � 0. We obviously do not want to - nor can - test all the numbers in these intervals. Thus, we only need to select one test point within each interval and see whether it satisfies the inequality. If this test point results in a true statement, then all values on this interval will satisfy the inequality. If the chosen 9-value does not satisfy the inequality, then no values on that interval will satisfy it.

Since we can select any number within each interval, let us choose test points�4, �1, and 3, which provide for easy calculations. The following chart displays the process.

Note: We can also use the factored form, 9 � 39 � 2 to test the numbers.

�4 � 3�4 � 2 = (�1)(�6) Positive sign

�1 � 3�1 � 2 = (2)(�3) Negative sign

3 � 33 � 2 = (6)(1) Positive sign

Observe that we only need to confirm if each of the resulting signs is negative or positive to check if the interval satisfies the inequality. We have seen that the expression 9� � 9 � 6 will be negative only on the interval �3, 2, so, the solution for the given inequality is �3, 2. The next graph confirms the solution values; the graph is below the x-axis at the interval �3, 2.

Interval Test Point Outcome for the Test Point

Resulting Sign Satisfies Inequality: < 0?

I: �∞,�3 �4 �4� ��4 � 6 + no

II: �3, 2 �1 �1� ��1 � 6 � yes

III: (2, ∞ 3 3� �3 � 6 + no

76

-4 -3 -2 -1 1 2 3 4

-6

-4

-2

2

4

6

x

y

Let us summarize the steps for solving a polynomial inequality by algebraic methods.

Solving a Polynomial Inequality Algebraically

(1) Write the given inequality in the formd9 < 0, d9 > 0, d9 ≤ 0, or d9 ≥ 0, as appropriate.

(2) Replace the inequality symbol with an equal sign and solve d9 = 0.

(3) Plot the solutions from (2) on a real number line. Use a solid dot if the inequality contains ≤ or ≥; if the inequality is < or >, use an open circle.

(4) Select one test point within each interval and see whether it satisfies the inequality.

(5) The solution will consist of the interval(s) that will satisfy the original inequality.

Example 4 Solve 90 � 89 ≥ 0 algebraically. Confirm your answer graphically.

Solution

Let us replace the inequality symbol with an equal sign and solve 90 � 89 " 0.

99� � 8 " 0 Factor (GCF).

99 � 2(9� − 29 + 4 = 0 Factor (sum of cubes).

9 = 0, or 9 + 2 = 0, or 9� − 29 + 4 = 0 Apply zero-product property.

Solving for x, we get 9 = 0 or 9 = −2. Notice that 9� − 29 + 4 = 0 has no real solution (the value of the discriminant is negative).

Now we plot these critical values on a real number line.

I II III

x

0 -2 (−∞,�2< [−2,0] [0, ∞

We will use test points�3, �1, and 1, which provide for easy calculations.

77

Hence, 90 � 89 ≥ 0 only on the first and third intervals, where the resulting sign is positive. The solution for the inequality is �∞,�2< ∪ [0, ∞, and the graph confirms these values.

-3 -2 -1 1 2

-8

-6

-4

-2

2

4

6

8

x

y

Example 5 Solve 9� � 39� − 9 − 3 < 0 algebraically. Confirm your answer graphically.

Solution

First, we replace the inequality symbol with an equal sign and solve for 9. Notice that the left-hand side can be factored by grouping.

9� + 39� − 9 − 3 = 0

(9� − 9+(39� − 3 = 0

9(9� − 1+3(9� − 1 = 0

(9+3(9� − 1 = 0

(9 + 3(9 + 1(9 − 1 = 0

Applying the zero-product property and solving for 9, we get 9 = −3, 9 = −1, and 9 = 1.

These critical values divide the real number line into four intervals.

I II III IV

x

0 -1 -3 1

(−∞,�3) (−3, �1 (−1,1 (1, ∞

We will use test points�4, �2, 0, and 2.

Interval Test Point Outcome for the Test Point

Resulting Sign Satisfies Inequality: ≥ 0?

I: �∞,�2< −3 (−30 + 8(−3 + yes

II: [−2, 0] �1 �10 + 8(−1 − no

III: [0, ∞ 1 10 + 8(1 + yes

78

From the sign chart, the solution for 9� � 39� − 9 − 3 < 0 is given by (−∞,�3 ∪ (−1,1. The graph confirms the solution.

-4 -3 -2 -1 1 2 3

-8

-6

-4

-2

2

4

6

8

x

y

Example 6 Kwana and Miriam built and successfully launched a model space rocket for the Physics Bowl at their university. The rocket was launched from ground level. The height in feet, ℎ, of the rocket after �seconds is modeled by the equation h = −16�2 + 132�. During what time interval was the rocket less than 182 feet high? Round your answer(s) to 2 decimal places. Solve algebraically and confirm your answer graphically.

Solution

We need to solve the following inequality:

−16�2 + 132� < 182

We will move all the nonzero terms to one side and solve the related equation.

−16�2 + 132� −182 = 0

The quadratic formula will help us solve the equation. The values for a, b, and c are: � = −16, � = 132, and = −182

Substituting these values into the quadratic formula, we get

� =(���±�(���+(0((�,((�-��((�,

Interval Test Point Outcome for the Test Point

Resulting Sign

Satisfies Inequality: < 0?

I: (−∞,�3 −4 (−4 + 3(−4 + 1(−4 − 1 − yes

II: (−3, �1 −2 (−2 + 3(−2 + 1(−2 − 1 + no

III: (−1,1 0 (0 + 3(0 + 1(0 − 1 − yes

IV: (1, ∞ 2 (2 + 3(2 + 1(2 − 1 + no

79

= (���±√�55,

(��

t ≈ 1.75 and t ≈ 6.5

Next, we plot these solutions on a real number line.

I II III

t

0 1.75 6.5

(∞, 1.75) (1.75, 6.5) (6.5, ∞)

Since � represents time, our model is relevant only for values of 0 seconds until the time the rocket falls to ground level, which we can find by solving �16�2 + 132� = 0. This gives us a landing time of � = 8.25 seconds. Therefore, we will test for the following intervals.

Thus, the rocket reached a height less than 182 feet during the intervals (0, 1.75) ∪ (6.5, 8.25). The graphical solution follows.

1 2 3 4 5 6 7 8

50

100

150

200

250

t

h

(1.75, 182) (6.5, 182)

Rational Inequalities

Let us begin by defining a rational inequality.

Rational Inequalities

A rational inequality is an inequality which can be written as

d9 < 0, d9 > 0, d9 ≤ 0, or d9 ≥ 0, where d9= i@t@is a rational function,

such that k9 and e9 are polynomials and e9 ≠ 0.

To solve a rational inequality, we can extend the graphing and algebraic techniques we used to solve polynomial inequalities.

Interval Test Point

Outcome for the Test Point

Resulting Sign Satisfies Inequality: < 0?

I: (0, 1.75) 0 �1602 + 1320 �182 � yes

II: (1.75, 6.5) 3 �1632 + 132(3 −182 + no

III: (6.5, 8.25) 7 −16(72 + 132(7 −182 − yes

80

Let us start with solving by graphing.

Example 7

The graph of d9= �@(0@(0 is given.

-6 -4 -2 2 4 6 8 10 12 14

-6

-4

-2

2

4

6

8

10

x

y

x = 4

Use the graph to solve the following.

a. �@(0@(0 ≥ 0 b. �@(0

@(0 < 0 Solution

a. The given function is not defined at 9 = 4, a it has a zero at 9 = 2. From the graph, we can see that the function will be on or above the x-axis when 9 ≤ 2 or 9 > 4. The solution in interval notation is given by (−∞,2< ∪ (4, ∞. b. The graph of the function lies below the 9-axis for 2 < 9 < 4, or the interval (2, 4.

Example 8

Solve the rational inequality by graphing: H(�HJ(� >0.

Solution

We graph d9= @(�@+(�. The function has vertical asymptotes at 9 = ±3, and a zero at 9 = 1.

-5 -4 -3 -2 -1 1 2 3 4 5

-6

-4

-2

2

4

6

x

y

x = -3 x = 3

The graph shows that the given function will be greater than zero for −3 < 9 < 1 or 9 > 3. The solution in interval notation is (−3, 1 ∪ (3, ∞.

81

Let us now solve rational inequalities algebraically.

Solving a Rational Inequality Algebraically

(1) Write the given inequality in the formd9 < 0, d9 > 0, d9 ≤ 0, or d9 ≥ 0, as appropriate. If there is more than one rational expression on the left-hand side of the inequality, combine them into a single fraction.

(2) Determine any value(s) where the numerator and the denominator are zero. These will be the critical values that will determine the intervals on the number line.

(3) Plot the solutions from (2) on a real number line. � For the numerator, use a solid dot if the inequality contains ≤ or ≥; if the inequality is < or >, use an open circle.

� Since we need to exclude any values that would make the denominator equal to zero on any rational expression in the inequality, we always use an open circle for critical values related to the denominator.

(4) Select one test point within each interval and see whether it satisfies the inequality.

(5) The solution will consist of the interval(s) that will satisfy the original inequality.

Example 9

Solve the rational inequality algebraically: @(�@+(� > 0.

Solution

We now follow the steps for solving a rational equation algebraically.

(1) The given function already satisfies this step.

(2) We will find the critical values by setting the numerator and denominator equal to zero, and solving for 9.

Numerator: 9 � 1 = 0 9 = 1 Denominator: (9 + 3(9 − 3 = 0 9 = −3 or 9 = 3

(3) These critical values divide the real number line into four intervals.

I II III IV

x

0 3 -3 1 (−∞,�3) (−3,1 (1, 3 (3, ∞

(4) We will construct the sign chart using test points�4, 0, 2, and 4.

82

(5) As we can see from the sign chart, the solution for @(�@+(� >0 is given by �3,1 ∪ (3, ∞,

which is the same solution we found solving the problem graphically in Example 8.

Example 10

Solve each the rational inequality algebraically. Confirm your answers graphically.

a. @

�@(� < ��@(� b.

@(�@&� ≤ @&�

@(0

Solution

a. (1) First, we will write the given inequality in the formd9 < 0 and combine the rational expressions as a single fraction.

@

�@(� � ��@(� < 0

@(��@(� < 0

(2) Next, we find the critical values by setting the numerator and denominator equal to zero, and solving for 9.

Numerator: 9 � 5 = 0 9 " 5 Denominator: 39 − 1 = 0 9 = 1/3

(3) In this example, the critical values divide the real number line into three intervals.

I II III

x

0 5 1/3 (−∞,1/3) (1/3, 5 5,∞ (4) We will use test points 0, 1, and 6.

Interval Test Point Outcome for the Test Point

Resulting Sign

Satisfies Inequality: > 0?

I: �∞,�3 −4 (−4 − 1(−4� − 9

− no

II: (−3,1 0 (0 − 1(0� − 9

+ yes

III: (1, 3 2 (2 − 1(2� − 9

− no

IV: (3, ∞ 4 4 � 1(4� − 9

+ yes

Interval

Test Point Outcome for the Test Point

Resulting Sign

Satisfies Inequality: < 0?

I: (−∞, 1/3) 0 0 − 53(0 − 1

+ no

II: (1/3, 5 1 1 − 53(1 − 1

− yes

III: (5, ∞ 6 6 � 53(6 − 1

+ no

83

(5) The solution of the given rational inequality is (1/3, 5). The next graph confirms the interval where the function is below the 9-axis.

-5 -4 -3 -2 -1 1 2 3 4 5 6 7

-6-5-4-3-2-1

123456

x

yx = 1/3

b. (1) Writing the given inequality in the formd9 ≤ 0 and combining the rational expressions as a single fraction, we have

@(�@&� � @&�

@(0 ≤ 0

@(�@(0@&�@(0 � @&�@&�

@(0@&� ≤ 0

F92−59+4G−(92+79+10

(@&�(@(0 ≤ 0

(,(29+1

(@&�(@(0 ≤ 0

(2) Now we find the critical values.

Numerator: −6(29 + 1 = 0 9 = −1/2 Denominator: (9 + 5(9 − 4 = 0 9 = −5 or 9 = 4

(3) The four intervals created by the critical values are shown next. I II III IV

x

0 4 -5 -1/2 (−∞,�5) �5,�1/2< [−1/2, 4 4, ∞ Note: Remember to use an open circle on �5 and 4. These values cannot be in the solution set because they make the denominator equal to 0.

(4) We will use test points�6,�1, 0, and 5.

Interval Test Point Outcome for the Test Point

Resulting Sign

Satisfies Inequality: ≤ 0?

I: �∞,�5 �6 �6[2(−6+ 1<(−6 + 5(−6 − 4

+ no

II: (−5, �1/2< −1 −6[2(−1+ 1<(−1 + 5(−1 − 4

− yes

III: [−1/2, 4 0 �6[2(0 + 1<(0 + 5(0 − 4

+ no

IV: (4, ∞ 5 �6[2(5 + 1<(5 + 5(5 − 4

− yes

84

(5) The solution of the given rational inequality is �5,�1/2< ∪ 4, ∞. The graph confirms the solution.

-12 -10 -8 -6 -4 -2 2 4 6 8 10 12

-6-5-4-3-2-1

12345

x

y

x = -5 x = 4 Example 11 Dulani has a successful business designing and selling sports shirts. He wants to create and sell sports shirts with special designs and donate all proceeds to a nonprofit organization dedicated to help fight Alzheimer's disease. The average cost per shirt, in dollars, for producing 9 number of

sports shirts is given by the function P9 = �..&0.5�@

@ . How many special sports designs shirts

can Dulani produce to maintain an average cost per unit of no more than $5? Solve algebraically and confirm graphically. Round your answer to the nearest whole number. Little Facts: Alzheimer's is a type of dementia that causes problems with memory, thinking and behavior. Its symptoms usually develop slowly worsen over time, to the point where it may interfere with daily tasks. Contrary to what may people believe, Alzheimer's is not just a disease of old age. Alzheimer's has no current cure, but treatments for symptoms are available and research continues. Source: www.alz.org.

Solution

We need to solve the following inequality: �..&0.5�@

@ � 5

Let us follow the steps for solving a rational equation algebraically.

(1) Writing the given inequality in the formd9 < 0 and combining the rational expressions as a single fraction, results in

�..&0.5�@

@ � 5)@@* � 0

�..&0.5�@(�@

@ � 0

�..(..��@

@ � 0

(2) The critical values are found next.

Numerator: 500 � 0.259 " 0 9 " 2000 Denominator: 9 " 0 9 " 0

85

(3) The critical values divide the real number line into three intervals.

I II III

x

0 2000 �∞,0) 0, 2000< ;2000, ∞

(4) Since 9 represents the number of sports shirts, our model is not relevant for negative values. So, we will test for the following intervals, using 100 and 2100.

(5) Observe from the sign chart that the solution is given by ;2000, ∞. Therefore, Dulani must produce a minimum of 2,000 shirts to maintain an average cost per unit of no more than $5.

The graph is shown below.

500 1000 1500 2000 2500 3000 3500

5

10

15

20

25

30

35

40

x

A(x)

xxxA 75.4500)( ++++====

In-Class Practice Section (G) 1. Solve the following inequalities by graphing.

a. 9� � 29� � 159 � 0 b. 90 � 29� ≥9� � 29

c. �@(0@+(�, �0 d.

@(0@&� � 6

2. Solve inequalities 1(a) and 1(c) algebraically. Confirm your answers with the graphs from problem 1.

3. The number of earthquakes in the United States from 2000-2010 can be modeled by the function f(x) = 37.105879� � 480.624139� � 1696.18959x + 1824.05594, where 9 is the number of years after 2000. Sources: www.earthquake.usgs.gov; www.mapsofworld.com

a. Graph the given function using the window [0, 10, 1] by [0, 9000, 1000]. b. Determine the years when there were more than 5000 earthquakes in the United States for the given period of time. Round your answer to the nearest year.

Interval Test Point Outcome for the Test Point

Resulting Sign

Satisfies Inequality: � 0?

II: 0, 2000< 100 500 � 0.25100

100

+ no

III: ;2000, ∞ 2100 500 � 0.252100

2100

� yes

86

Exercises Section (G) In exercises 1-9, solve the following inequalities algebraically and write your answers using interval notation. Confirm your answers graphically. Write answers in fraction form where needed; do not use decimals.

1. 0252523 >−−+ xxx 2. 02446 23 ≤−++− xxx 3. 032162 234 ≥−++− xxxx 4. 04950 24 >+− xx 5. 0214 23 <−− xxx 6. 01572 23 ≥−+ xxx 7. 182723 23 −≤− xxx 8. 3613 24 −< xx 9. 445 42 −−≥− xx In exercises 10-15, solve the following inequalities using your graphing calculator. Write your answers in interval notation and round to 2 decimal places where needed. 10. 20118 23 ≤++ xxx 11. 423 12112 xxxx −−>−− 12. 324 21632 xxxx −−>− 13. 75.375.73 −≤+− xx 14. 06.229.33 234 ≥+−− xxxx 15. 02.06.52 24 ≥−+− xx In exercises 16-22, solve each the rational inequality algebraically. Write your answers in interval notation.

16. 049

22

<−+

x

x 17. 0

36

52

≥−+

x

x 18. 0

2110

12

≥++

−xx

x 19. 0

4

522

≤−−

xx

x

20. 12

4

12

7

−<

−+

xx

x 21.

1

3

7 −≤

+ xx

x 22.

x

x

x

x 4

3

6 +≤−−

23. 10

2

8

1

−≥

−+

x

x

x

x

24. Sharma teaches a Peace Studies course at a university in Delhi, India. Her course focuses around the teachings of Mahatma Gandhi. The number of students enrolled in her course over a 5 year period can be modeled by the function �� = 9.09�0 − 71.4�� + 170.9�� − 108� + 50, where �represents the year and �(� represents the number of students. Graph the function and determine the years in which there were at least 120 students enrolled. Write your answer in interval notation, rounding to 2 decimal places. Little Facts: Mohandas Gandhi was born on October 2, 1869 in Gujarat, India. He left India to study law in London at the age of 19. Soon after, Gandhi and his family moved to South Africa; though they spent 20 years there, they faced fierce discrimination. After being beat up for refusing to give up his seat to a European, Gandhi began his journey to free others due to discrimination through passive resistance. In 1914, they returned to India where he became known as the father of the Indian independence movement, hence given the name Mahatma which translates into “Great Soul." Gandhi was assassinated in 1948 at the age of 78. Sources: www.biography.com; www.history1990s.about.com 25. The amount of sales of a local organic grower can be modeled by the function f(9 = −0.00394 + 0.11893 − 1.292 + 2.549 + 22, where f(9 is sales in thousands and 9 represents months over a 25 month period. Graph the function and determine the months in which their sales were over $27,000. Write your answer in interval notation, rounding to 2 decimal places.

87

26. Lynette is a seamstress who decides to make stylish versions of night robes. The average cost per night robe, in dollars, for producing 9 number of robes is given by the function

P9 = ,..&5.5�@

@ . How many night robes must she sew to maintain an average cost per

robe of no more than $12? Solve algebraically and confirm graphically. Round your answer to the nearest whole number.

Answer Key Section (G) Exercises 1. ),5()1,5( ∞∪−− 3. ]4,2[]0,4[ ∪− 5. )7,0()3,( ∪−−∞ 7. ]3,3/2[]3,( ∪−−∞

9. ),( ∞−∞ 11. ),4()1,0()3,( ∞∪∪−−∞ 13. ),3[]5.0,5.2[ ∞∪−−

15. ]66.1,19.0[]19.0,66.1[ ∪−− 17. ),6(]5,6( ∞∪−− 19. )4,2/5[)0,( ∪−∞

21. �7, �3< ∪ (1,7< 23. ]10,8(]5,2[ ∪

25. (13.25, 24.04)

4 8 12 16 20 24 28

10

20

30

40

50

60

x

S(x)

∪