MTH 209 Week 4 Third. Final Exam logistics Here is what I've found out about the final exam in...

MTH 209 Week 4

Third

Final Exam logistics

Here is what I've found out about the final exam in MyMathLab  (running from the end of class this week (week 4 at 10pm)  to 11:59pm five days after the last day of class.

.

Slide 2Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Final Exam logistics There will be 50 questions. You have only one attempt to complete the exam. Once you start the exam, it must be completed in that sitting.   (Don't start until you have

time to complete it that day or evening.) You may skip and get back to a question BUT return to it before you hit submit. 

You must be in the same session to return to a question. There is no time limit to the exam (except for 11:59pm five nights after the last class). You will not have the following help that exists in homework:

Online sections of the textbook Animated help Step-by-step instructions Video explanations Links to similar exercises

You will be logged out of the exam automatically after 3 hours of inactivity.  Your session will end.

IMPORTANT!  You will also be logged out of the exam if you use your back button on your browser.  You session will end.

Slide 3Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Section 6.6

Solving Equations by

Factoring I (Quadratics)

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• The Zero-Product Property

• Solving Quadratic Equations

• Applications

Zero-Product Property

To solve equations we often use the zero-product property, which states that if the product of two numbers is 0, then at least one of the numbers must be 0.

Example

Solve each equation.a. b. Solutiona. b.

3 ( 4) 0x x (4 1)(3 4) 0x x

3 ( 4) 0x x 04( )3x x

3 0 or 4 0 x x

0 o r 4 x x

(4 1)(3 4) 0x x 3( 4 01)(4 )xx 0 or 31 044 xx

4 1 or 3 4x x 1 4

or 4 3

x x

Try Q’s pg 402 13,15,19,23

Solving Quadratic Equations

Any quadratic polynomial in the variable x can be written as ax2 + bx + c with a ≠ 0.

Any quadratic equation in the variable x can be written as ax2 + bx + c = 0, with a ≠ 0. This form of quadratic equation is called the standard form of a quadratic equation.

Slide 9

Example

Solve each quadratic equation. Check your answers.a. 36 + x2 = –12x b. x2 − 49 = 0Solutiona. 36 + x2 = –12x

x2 + 12x + 36 = 0(x + 6)(x + 6) = 0

x = −6

The only solution is −6.

To check this value, substitute −6 for x in the given equation.

36 + x2 = –12x

36 + (−6)2 = –12(−6)

72 = 72

Example (cont)

b. x2 − 49 = 0(x + 7)(x − 7) = 0

x =

−7

x + 7 = 0 x − 7 = 0x = 7

To check these values, substitute 7 and −7 for x in the given equation.

72 − 49 = 00 = 0

(−7)2 − 49 = 0 0 = 0

The solutions are −7 and 7.

Try Q’s pg 403 27,35,43,49

Example

Solve 2x2 − 7x = −5Solution

The solutions are

(2x – 5)(x − 1) = 0

2x – 5 = 0 or x – 1 = 0

or x = 1

2x2 − 7x = −52x2 − 7x + 5 = 0

5

2x

51 and .

2

Try Q’s pg 403 51

Example

If a model rocket is launched at 48 feet per second, then its height, h, after t seconds is h = 48t – 16t2. After how long does the rocket strike the ground?SolutionThe rocket strikes the ground when the height is 0.

48t – 16t2 = 0

16t(3 – t) = 0

t = 0

16t = 0 3 – t = 0

t = 3

The rocket strikes the ground after 3 seconds.

Try Q’s pg 403 67a

Example

A frame surrounding a picture is 2 inches wide. The picture inside the frame is 7 inches longer than it is wide. If the overall area of the picture and frame is 198 square inches, find the dimensions of the picture inside the frame.SolutionLet x be the width of the picture and x + 7 be its length.

2

2

2

2

x + 7

x

x + 11

x + 4

Example (cont)

(x + 4)(x + 11) =198

x2 + 15x + 44 = 198

x2 + 15x − 154 = 0

(x − 7)(x + 22) = 0

x − 7 = 0 or x + 22 = 0

x = 7 or x = −22

The only valid solution for x is 7 inches. Because the length is 7 inches more than the width, the dimensions are 7 inches and 14 inches.

Try Q’s pg 404 73

Section 8.4

Other Functions and Their Properties

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Expressing Domain and Range in Interval Notation

• Absolute Value Function

• Polynomial Functions

• Rational Functions (Optional)

• Operations on Functions

Expressing Domain and Range in Interval Notation

The set of all valid inputs for a function is called the domain, and the set of all outputs from a function is called the range.

Rather than writing “the set of all real numbers” for the domain of f, we can use interval notation to express the domain as (−∞, ∞).

Example

Write the domain for each function in interval notation.

a. f(x) = 3x b.Solutiona. The expression 3x is defined for all real numbers x. Thus the domain of f is

b. The expression is defined except when x – 4 = 0 or x = 4. Thus the domain of f includes all real numbers except 4 and can be written

1

4f x

x

, .

, 4 4, .

Try Q’s pg 563 13,21,25

Absolute Value Function

We can define the absolute value function by f(x) = |x|.

To graph y = |x|, we begin by making a table of values. x |x|

–2 2

–1 1

0 0

1 1

2 2

Example

Sketch the graph of f(x) = |x – 3|. Write its domain and range in interval notation. SolutionStart by making a table of values.

x y

0 3

2 1

3 0

4 1

6 3

X

Y

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

-5

-4

-3

-2

-1

1

2

3

4

5

0

The domain of f is , .

The range of f is [0, ).

Try Q’s pg 563 39

Polynomial Functions

The following expressions are examples of polynomials of one variable.

As a result, we say that the following are symbolic representations of polynomial functions of one variable.

2 3, , and 51 515 3x xx x

32( ) 3, , and ( 5 1 ( )) 1 5 5f g x x xx xx hx

Example

Determine whether f(x) represents a polynomial function. If possible, identify the type of polynomial function and its degree. a.

b.

c.

3( ) 6 2 7f x x x

3.5( ) 4f x x

4( )

5f x

x

cubic polynomial, of degree 3

not a polynomial function because the exponent on the variable is negative

not a polynomial

Try Q’s pg 564 43,45,47,51

Example

A graph of is shown. Evaluate f(1) graphically and check your result symbolically.Solution

3( ) 5f x x x

To calculate f(–1) graphically find –1 on the x-axis and move down until the graph of f is reached. Then move horizontally to the y-axis.

f(1) = –4 3( 1) 5( ( )1 1)f 5 1

4

Try Q’s pg 564 59,73

Example

Evaluate f(x) at the given value of x.

Solution

3 2( ) 4 3 7, 2f x x x x

3 2( ) 4 3 7, 2f x x x x

3 2( 2) 4( 2) 3( 2) 7f

( 2) 4( 8) 3(4) 7f

( 2) 32 12 7f

( 2) 27f

Try Q’s pg 564 55-64

Example

Use and to evaluate each of the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

2a. ( ) 3 11 (1)

4

f

2( ) 61 ( )

5

1g

( )(1) (1) (1)

4 5

9

f g f g

Example (cont)

Use and to evaluate each of the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

2b. ( ) 3( ) 1

3(4

2

13

2

) 1

f 2( ) 6 (

6 4

2

2 )2g ( )( 2) ( 2) ( 2)

13 2

11

f g f g

Example (cont)

Use and to evaluate each of the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

c. 0f

g

0

00

ff

g g

2

2

3( ) 1 =

6 )

0

(0

1 =

6

Try Q’s pg 564 64-70

Section 11.1

Quadratic Functions and Their Graphs

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Graphs of Quadratic Functions

• Min-Max Applications

• Basic Transformations of Graphs

• More About Graphing Quadratic Functions (Optional)

The graph of any quadratic function is a parabola.

The vertex is the lowest point on the graph of a parabola that opens upward and the highest point on the graph of a parabola that opens downward.

The graph is symmetric with respect to the y-axis. In this case the y-axis is the axis of symmetry for the graph.

Example

Use the graph of the quadratic function to identify the vertex, axis of symmetry, and whether the parabola opens upward or downward.a. b.

Vertex (0, 2)Axis of symmetry: x = –2Open: up

Vertex (0, 4)Axis of symmetry: x = 0Open: down

Try Q’s pg 725 29,31

Example

Find the vertex for the graph of Support your answer graphically.Solutiona = 2 and b = 8

Substitute into the equation to find the y-value.

2( ) 2 8 3.f x x x

82

2(2)x

2

bx

a

2( ) 2( 2) 8( 2) 3

8 16 3

11

f x

The vertex is (2, 11), which is supported by the graph.

Try Q’s pg 724 15

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (0, –2)axis of symmetry x = 0

2( ) 2f x x

x f(x) = x2 – 2

3 7

2 2

1 1

0 2

1 1

2 2

3 7

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (2, 0)axis of symmetry x = 2

2( ) ( 2)g x x

x g(x) = (x – 2)2

0 4

1 1

2 0

3 1

4 4

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (2, 0)axis of symmetry x = 2

2( ) 2 3h x x x

x h(x) = x2 – 2x – 3

2 5

1 0

0 3

1 4

2 3

3 0

4 5

Example

Find the maximum y-value of the graph of

SolutionThe graph is a parabola that opens downward because a < 0. The highest point on the graph is the vertex.a = 1 and b = 2

2( ) 2 3.f x x x

( 2)1

2 2( 1)

bx

a

2( ) ( 1) 2( 1) 3

4

f x

Try Q’s pg 726 81

Example

A baseball is hit into the air and its height h in feet after t seconds can be calculated by a. What is the height of the baseball when it is hit?b. Determine the maximum height of the baseball.Solutiona. The baseball is hit when t = 0.

b. The graph opens downward because a < 0. The maximum height occurs at the vertex. a = –16 and b = 64.

2( ) 16 64 2.h t t t

2( ) 16 64 2h t t t 2(0) 16(0) 64(0) 2

2

h

64 642

2 2( 16) 32

bx

a

Example (cont)

2( ) 16 64 2 h t t t

2(2) 16(2) 64(2) 2

66

h

The maximum height is 66 feet.

2( ) 16 64 2 h t t t

Try Q’s pg 726 87

Basic Transformations of Graphs

The graph of y = ax2, a > 0.As a increases, the resulting parabola becomes narrower.When a > 0, the graph of y = ax2 never lies below thex-axis.

Section 11.2

Parabolas and Modeling

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Vertical and Horizontal Translations

• Vertex Form

• Modeling with Quadratic Functions (Optional)

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).

All three graphs have the same shape.y = x2

y = x2 + 1 shifted upward 1 unity = x2 – 2 shifted downward 2 units

Such shifts are called translations because they do not change the shape of the graph only its position

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2

y = (x – 1)2 Horizontal shift to the right 1 unit

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2

y = (x + 2)2 Horizontal shift to the left 2 units

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated 3 units down.

The vertex is (0, 3).

2 3y x

Try Q’s pg 739 15,19,27

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated left 4 units.

The vertex is (4, 0).

2( 4)y x

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated down 2 units and right 1 unit.

The vertex is (1, 2).

2( 1) 2y x

Try Q’s pg 739 39,37

Example

Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane.

SolutionThe graph is translated to the right 2 units and upward 3 units.The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0).The graph opens upward and is wider.

21( ) ( 2) 3

4f x x

Try Q’s pg 739 41

Example

Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c.SolutionThe vertex form of the parabola is where the vertex is (h, k).a = 3, h = 2 and k = 1

To write the equation in y = ax2 + bx + c, do the following:

2( ) ,y a x h k

2)3( 12y x

2)3( 12y x 2( 4 43 1)y x x

23 12 12 1y x x 23 12 13y x x

Try Q’s pg 740 65,68

Example

Write each equation in vertex form. Identify the vertex. a. b.Solutiona. Because , add and subtract 16

on the right.

2 8 13y x x 22 8 7y x x 2 2

816

2 2

b

2 8 13y x x

2 16 1 68 13y x x

24 3y x

The vertex is (4, 3).

Example (cont)

b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side.

22 8 7y x x

2 24

42 2

b

2

2

2 8 7

2( 4 ) 7

y x x

x x

2 42 4 74y x x

2 42 4 7 8y x x

22 2 1y x The vertex is ( 2, 1).

Try Q’s pg 739 53

Section 11.3

Quadratic Equations

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Basics of Quadratic Equations

• The Square Root Property

• Completing the Square

• Solving an Equation for a Variable

• Applications of Quadratic Equations

Basics of Quadratic Equations

Any quadratic function f can be represented by f(x) = ax2 + bx + c with a 0.

Examples:

2 2 21( ) 2 1, ( ) 2 , and ( ) 2 1

3f x x g x x x h x x x

Basics

The different types of solutions to a quadratic equation.

Example

Solve each quadratic equation. Support your results numerically and graphically. a. b. c. Solutiona. Symbolic: Numerical: Graphical:

23 2 0x 2 9 6x x 2 2 8 0x x

The equation has no real solutions because x2 ≥ 0 for all real numbers x.

x y

1 5

0 2

1 5

2

2

2

3 2 0

3 2

2

3

x

x

x

Example (cont)

b. 2 9 6x x 2

2

9 6

6 9 0

( 3)( 3) 0

3 0 or 3 0

3 or 3

x x

x x

x x

x x

x x

The equation has one real solution.

x y

5 4

4 1

3 0

2 1

1 4

Example (cont)

c. 2 2 8 0x x 2 2 8 0

( 2)( 4) 0

2 0 or 4 0

2 or 4

x x

x x

x x

x x

The equation has two real solutions.

x y

4 0

2 8

1 9

0 8

2 0

Try Q’s pg 752 29,37,39

The Square Root Property

The square root property is used to solve quadratic equations that have no x-terms.

Example

Solve each equation. a. b. c.Solutiona.

2 10x 225 16 0x 2( 3) 16x

2 10x

2 10x

10x

b. 225 16 0x 225 16x

2 16

25x

16

25x

4

5x

c. 2( 3) 16x

( 3) 16x

3 4x

3 4x

1 or 7x

Try Q’s pg 752 51,53,57

Real World Connection

If an object is dropped from a height of h feet, its distance d above the ground after t seconds is given by

2( ) 16d t h t

Example

Find the term that should be added to to form a perfect square trinomial.

SolutionCoefficient of x-term is –8, so we let b = –8. To complete the square we divide by 2 and then square the result.

2 8x x

2 28

162 2

b

2 2168 ( 4)x x x

Try Q’s pg 752 67

Example

Solve the equationSolutionWrite the equation in x2 + bx = d form.

2 8 13 0x x

2 28

162 2

b

2 8 13x x 2 168 3 61 1x x

2( 4) 3x

4 3x

4 3x

5.73 or 2.27x

Try Q’s pg 752 73

Example

Solve the equationSolutionWrite the equation in x2 + bx = d form.

20 2 8 7x x

2 24

42 2

b

22 8 7x x

2 442

47

x x

2 1( 2)

2x

12

2x

2 74

2x x

22

2x

12

2x

22

2x

1.29 or 2.71x

Try Q’s pg 752 81

Example

Solve the equation for the specified variable.Solution 216 for w x x

216w x

16

wx

4

wx

2

16

wx

Try Q’s pg 753 105,107

Section 11.4

Quadratic Formula

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Solving Quadratic Equations

• The Discriminant

• Quadratic Equations Having Complex Solutions

, 0 and 1,xf x a a a

The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by

QUADRATIC FORMULA

2 4.

2

b b acx

a

Example

Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically.SolutionSymbolic SolutionLet a = 4, b = 3 and c = − 8.

2 4

2

b b acx

a

23 3 4 4 8

2 4x

3 137

8x

3 137

8x

or 3 137

8x

1.1x 1.8x or

Example (cont)

4x2 + 3x – 8 = 0Graphical Solution

Try Q’s pg 764 9

Example

Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically.SolutionLet a = 3, b = −6 and c = 3.

2 4

2

b b acx

a

26 6 4 3 3

2 3x

6 0

6x

1x

Try Q’s pg 764 11

Example

Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically.SolutionLet a = 2, b = 4 and c = 5.

2 4

2

b b acx

a

24 4 4 2 5

2 2x

4 24

4x

There are no real solutions

for this equation because

is not a real number.

24

Try Q’s pg 764 13

, 0 and 1,xf x a a a

To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac.

1. If b2 – 4ac > 0, there are two real solutions.

2. If b2 – 4ac = 0, there is one real solution.

3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.

THE DISCRIMINANT AND QUADRATIC

EQUATIONS

Example

Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula.Solution−2x2 + 5x − 3 = 0Let a = −2, b = 5 and c = −3.

Thus, there are two solutions.

b2 – 4ac

= (5)2 – 4(−2)(−3) = 1

2 4

2

b b acx

a

5 1

2 2x

4

4x

1x

or

6

4x

1.5x

Try Q’s pg 764 39a, b

, 0 and 1,xf x a a a

If k > 0, the solution to x2 + k = 0 are given by

THE EQUATION x2 + k = 0

.x i k

Example

Solve x2 + 17 = 0.

SolutionThe solutions are

17.i

17 or 17.x i i

Try Q’s pg 764 57

Example

Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. SolutionLet a = 3, b = −7 and c = 5. 2 4

2

b b acx

a

7 11

6x

and

27 7 4 3 5

2 3x

7 11

6

ix

7 11

6 6x i

7 11

6 6x i

Try Q’s pg 765 73

Example

Solve Write your answer in standard form: a + bi. SolutionBegin by adding 2x to each side of the equation and then multiply by 5 to clear fractions.

Let a = −2, b = 10 and c = −15.

223 2 .

5 x

x

22 10 15 0x x

Example (cont)

Let a = −2, b = 10 and c = −15.

223 2 .

5 x

x

2 4

2

b b acx

a

10 20

4x

210 10 4 2 15

2 2x

10 2 5

4

ix

5 5

2 2x i

Try Q’s pg 765 79

Example

Solve by completing the square.SolutionAfter applying the distributive property, the equation becomes

Since b = −4 ,add to each side of the equation.

4 5x x

2 4 5.x x

242 4

2 4 4 5 4x x 2

2 1x 2 1x 2x i

2x i

The solutions are 2 + i and 2 − i.

Try Q’s pg 765 91

Section 11.5

Quadratic Inequalities

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Basic Concepts

• Graphical and Numerical Solutions

• Symbolic Solutions

Example

Determine whether the inequality is quadratic.a. 6x + 2x2 – x3 ≥ 0 b. 8 + 7x2 + 2 < 6x2 + xSolutiona. The inequality 6x + 2x2 – x3 ≥ 0 is not quadratic

because it has an x3-term.

b. Write the inequality as follows. 8 + 7x2 + 2 < 6x2 + x

10 + 7x2 < 6x2 + x

x2 – x + 10 < 0

The inequality is quadratic because it can be written in the form ax2 + bx + c > 0 with a = 1, b = −1, and c = 10.

Try Q’s pg 775 7,11

Example

Make a table of values for x2 − 2x – 15 and then sketch the graph. Use the table and graph to solve x2 – 2x – 15 ≤ 0. Write your answer in interval notation.Solution

Interval notation [−3, 5]

x y = x2 – 2x – 15

−3 0

−2 −7

−1 −12

0 −15

1 −16

2 −15

3 −12

4 −7

5 0

Example (cont)

x2 − 2x – 15

Try Q’s pg 775 19,27

Example

Solve x2 > 4. Write your answer in interval notation.SolutionThe graph of y = x2 – 4 is shown with intercepts −2 and 2. Thus the solution set is given by x < −2 or x > 2, which can be written in interval notation as

, 2 2, .

Try Q’s pg 776 49

Example

Solve each of the inequalities graphically.a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0Solution a. Because the graph is always

above the x-axis, x2 + 3 is always greater than 0. The solution set includes all real numbers, or (−∞,∞).

b. Because the graph never goes below the x-axis, x2 + 3 is never less than 0. Thus there are no real solutions.

Example (cont)

Solve each of the inequalities graphically.a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0Solution a. Because the graph never

goes below the x-axis, (x − 3)2 ≤ 0 is never less than 0. When x = 3, y = 0, so 3 is the only solution to the inequality (x − 3)2 ≤ 0.

Try Q’s pg 776 51,55,57

Example

Solve each inequality symbolically. Write your answer in interval notation. a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9xSolutiona. x2 – 4x – 12 = 0

(x + 2)(x – 6) = 0

x = −2

x + 2 = 0 x – 6 = 0x = 6

The solutions lie outside the values. In interval notation the solution set is( , 2] [6, ).

Example (cont)

Solve each inequality symbolically. Write your answer in interval notation. a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9xSolutionb. – x2 − 9x + 10 < 0

– x2 − 9x + 10 = 0

(x + 10)(x – 1) = 0

x + 10 = 0 x – 1 = 0

x = −10 x = 1

The solutions lie between these two values. In interval notation the solution set is (−10, 1).

x2 + 9x − 10 = 0

Try Q’s pg 776 31,37

Example

A rectangular building needs to be 9 feet longer than it is wide. The area of the building must be at least 532 square feet. What widths x are possible for this building?Solution

x2 + 9x = 532

x(x + 9) ≥ 532

x2 + 9x – 532 = 0

29 9 4 1 532

2 1x

29 9 4 1 532

2 1

9 2209

2

9 47

2

28, 19

The width is positive, so the building must be 19 feet or more, x ≥ 19 feet.

Try Q’s pg 776 67

Due for this week…

Homework 4 (on MyMathLab – via the Materials Link) The fifth night after class at 11:59pm.

Read Chapter 12.1, 12.3 and 14.1-14.3 Do the MyMathLab Self-Check for week 4. Learning team hardest problem assignment. Complete the Week 4 study plan after submitting

week 4 homework. Participate in the Chat Discussions in the OLS

Slide 104Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

End of week 4 You again have the answers to those problems not assigned Practice is SOOO important in this course. Work as much as you can with MyMathLab, the materials

in the text, and on my Webpage. Do everything you can scrape time up for, first the hardest

topics then the easiest. You are building a skill like typing, skiing, playing a game,

solving puzzles. Next Week: Composite Functions, Inverse Functions,

Logarithmic functions, Arithmetic/Geometric sequences/series.