http://prezi.com/7celq_oomduu/algebra-two-section-54/

5.4 Prezi – By Anna

Chapter 5.5By Corey Lorraine

Warm Up Question

Vocabulary

Example 1

Example 2

Example 3

Example 3

Quadratic Formula and The Discriminant

Chapter 5 Section 6By, Tim Summers

Quadratic Formula X= (-b±√(b^2-4ac))/2a This is used to find the solution(s) of the

quadratic equation ax^2 + bx + c. Before using the quadratic equation ax^2 +

bx + c must be equal to zero. To check your answer replace the 0 with a

“y” and then insert the equation into a graphing calculator

Discriminant The discriminant is b^2 – 4ac, where the a,

b, and c are coefficients. The discriminant is used to tell how many

solutions and what type of solutions there are for a quadratic equation.

Discriminant If b^2 – 4ac > 0, then there are two real

solutions. If b^2 – 4ac = 0, then there is one real

solution. If b^2 – 4ac < 0, then there are two

imaginary solutions.

Example Problem #1

5x^2 + 6x + 1 = 0a = 5, b = 6, c = 1x = (-6 ± √(6^2 - 4×5×1)) / (2×5)x = (-6 ± √(36 - 20)) / 10x = (-6 ± √(16)) / 10x = (-6 ± 4) / 10x = (-6 – 4) / 10 or x = (-6 + 4) / 10x = -1 or x = -0.2

Example Problem #2

5x^2 + 2x + 1 = 0a = 5, b = 2, c = 1x = (-2 ± √(2^2 - 4×5×1)) / (2×5)x = (-2 ± √(4 - 20)) / 10x = (-2 ± √(-16)) / 10x = (-2 ± 4i) / 10x = -0.2 ± 0.4i

Example Problem #3

x^2 + 3x – 4 = 0a = 1, b = 3, c = -4x = (-3 ± √(3^2 - 4×1×(-4))) / (2×1)x = (-3 ± √(9 – (-16))) / 2x = (-3 ± √(25)) / 2x = (-3 ± 5) / 2x = (-3 + 5) / 2 or x = (-3 - 5) / 2x = 1 or x = -4

Section: 5.7Graphing and Solving

Quadratic InequalitiesBy: Andrew Fratoni

Quadratic Inequalities

Quadratic inequality in two variables

Quadratic inequality in one variable

Graphing a Quadratic InequalityStep: 1 Make the necessary parabola using the

equation y= ax2+ bx+ c

Step: 2 Make it with a

dashed line for inequalities using < or > and a solid line for inequalities using ≤ or ≥.

y > x2 + 3x – 4

y < 2x2 – 3x + 1

Step: 3 Choose a point either

outside or inside of the parabola and check to see if the point is a solution to the inequality

If it is a solution, then shade the corresponding area. If not, then shade the other side of the parabola

Graph y ≤ -2x2

Step 1: graph y = -2x2 Step 2: since the

inequality symbol is ≤, make the line solid

Step 3: test a point like (0,2)

2 ≤ -2(0) 2

2 ≤ 0 false. Shade the inside of the parabola

Graph y < x2 – 4x + 1

Step 1: graph y = x2 – 4x + 1

Step 2: since the inequality symbol is <, make the line dotted

Step 3: test a point like (2,0)

0 < (2)2 – 4(2) + 10 < -3 falseShade the area outside

of the parabola.

Graph the system of inequalities: y ≤ - x2 + 3 y ≥ x2 + 2x - 4

y ≤ - x2 + 3 Step 1: graph y = - x2 + 3Step 2: since the inequality

symbol is ≤, make the line solid

Step 3: test a point like (0,0)0 ≤ - (0)2 + 3 0 ≤ 3 True

y ≥ x2 + 2x – 4Step 1: graph y = x2 + 2x – 4Step 2: since the inequality

symbol is ≥, make the line solid

Step 3: test a point like (0,0)0 ≥ (0)2 + 2(0) – 40 ≥ -4 True

The shaded region is the area where both parabolas would be shaded and that satisfy both inequalities.

Solving Quadratic Inequalities AlgebraicallyFirst, replace the inequality

symbol with an equals sign.

Then, solve for x using one of the methods (quad. formula, completing the square, factoring, etc.)

Last, test an x-value in-between the two x values found above. Figure out if it satisfies the inequality. Write the solution based on the known information.

Ex. X2 + 3x – 18 ≥ 0X2 + 3x – 18 = 0 replace w/

equals sign

(x + 6)(x -3) = 0 factor

x = -6 x = 3 Set both equal to zero

Test x = 0(0)2 + 3(0) – 18 ≥ 0

-18 ≥ 0 falseSo, because a number between

the two values does not satisfy the inequality

Solution: x ≤ -6 or x ≥ 3

Real life Application

Finding the weight of theater equipment that a rope can support.

Ex. Weight that the rope can safely support (W) with diameter (d)

W≤ 1480d2

Using Propertiesof Exponents

6.1

PRODUCT OF POWERS PROPERTY am . an = am+n

POWER OF A POWER PROPERTY (am ) n = amn

POWER OF A PRODUCT PROPERTY (ab) m = am bm

NEGATIVE EXPONENT PROPERTY a-m= 1 / (am1)

ZERO EXPONENT PROPERTY a0 = 1, a 0 0

QUOTIENT OF POWERS PROPERTY am/ an = am-n, 0

POWER OF A QUOTIENT PROPERTY (a/b)m = am /bm 0

Properties of Exponents

Evaluating Numerical Expressions

215= 32,768

(32/42) = 9/16

(4) -3= 1/4 3 = 1/64(4)3. (4)-6 =?

(3/4)2 =?

(25)3 = ?

1.) 52 . 52

2.) (3-3)4

3.) (2/4)3

4.) 50.63

Student Practice1.) 625

2.) 1/531,441

3.) 1/8

4.) 108

Properties of Rational

Exponents(7.2)

PRODUCT OF POWERS PROPERTY am . an = am+n

POWER OF A POWER PROPERTY (am ) n = amn

POWER OF A PRODUCT PROPERTY (ab) m = am bm

NEGATIVE EXPONENT PROPERTY a-m= 1 / (am1)

ZERO EXPONENT PROPERTY a0 = 1, a 0 0

QUOTIENT OF POWERS PROPERTY am/ an = am-n, 0

POWER OF A QUOTIENT PROPERTY (a/b)m = am /bm 0

Properties of Rational Exponents

1.) n√a . b = ?

2.) n(√a/b) = ?

Student P rac t i ce

1.) n√a . n√b

2.) n√a/ n√b

6.2 evaluating and graphing polynomial function

ALGEBRA 2

when a0 is not zero, the exponents will be all whole numbers, and the coefficients will be all real numbers. for this polynomial function, the leading coefficient will be an, constant will be a0, and the degree will be na standard form polynomial function is when the terms of exponents are written in descending order and it also from left to right.

a polynomial function is a function of the form 1

1 2 1( ) n nn na x a x a x a x a

degree type stand form

0 constant F(x)=a0

1 linear F(x)=a1x+a0

2 Quadratic

3 cubic

4 quartic

Here is a summary of common types of polynomial function

Solution

coefficients

polynomial in standard form

3 0 -8 5 -7

4

3 12 40 165 653

12 48 160 660

f(4)=653

  x-value

Example 1use synthetic substitution to evaluate f(x)=38+5x-7 when x=4

0.000025 -0.004 0.33 5.3

100 0.0025 -0.21 12

0.000025 -0.0021 0.12 17.3

the recharge time is about 17 seconds

Example 2find the recharge time  of battery after 100 flashes, if the modeled is  where t(in sec) is  t=0.000025+0.004+0.33n+5.3.the time of a battery to recharge after flashing n times of uses.

Graph function

f (x)= x f (x)=-x

f (x)→+∞ as x →+∞

f (x)→-∞ as x →-∞

f (x)→+∞ as x →-∞

f (x)→-∞ as x →+∞

f (x)=x

f (x)=-x

f (x)→ +∞

As x→ -∞

f (x)→ +∞As x→ -∞

f (x)→ -∞As x→ -∞

f (x)→ -∞As x→ +∞

 

Solutiona. To graph this function, make a table of values and plot with corresponding points. Connect the points with a smooth curve and check the end behavior.

The degree is odd and the leading coefficient is positive so

f(x) → -∞ as x →-∞ and f(x) → +∞ as x →+∞

0

0-1-2-3X 321

f(x)

0-1-2-3X 321

-10-3-16

0-1-2-3X

329

321

solutionb. To graph this function, make a table of values and plot with corresponding points. Connect the points with a smooth curve and check the end behavior

18-15

f(x)

-1-2-3x

-32

3210

0 1 -147

The degree is even and the leading coefficient is negative so

f(x) → -∞ as x →-∞ and f(x) → -∞ as x →+∞

 

The end

Algebra IILesson 6.3

Adding,Subtracting, Multiplying Polynomials

J.W.Tang

~Adding Polynomials~

To add polynomials Combine like terms(make sure all degrees are

account for!) You can either do this vertically or horizontally

Example (pg 341 #25)

(10X-3+7X²)+(X³-2X+17) → =( X³+7X²+8X+14)

→ ___7X²+10X-3

+ X³ ___-2X+17-------------------------------

X³+7X²+8X+14

~Subtracting Polynomials~

To subtract polynomials Subtract like terms To make it easier flip signs :

multiply the equation you are subtracting by -1 and then add the equations

Example(pg341 #23)

(10X³-4X²+3X)-(X³-X²+1)

→ 10X³-4X²+3X__ 10X³-4X²+3X__

- X³ - X² __ +1 → + -X³ + X² __ -1

------------------- ------------------------

9X³-3X²+3X-1 9X³-3X²+3X+1

~Multiplying Polynomials~

To multiply polynomials set the equation like a normal multiplication equation then multiply one by one like a normal multiplication problem

See Next 2 Slides for Examples on how to multiply polynomials vertically and horizontally

-To multiply Vertically-Example (pg.341 #41)

(3X²-2) (X²+4X+3)

X²+4X+3

* 3X²-2

------------------

-2x²-8X-6

+ 3X412x³+9x²__

________________

3X412x³+7X²-8X-6

-To Multiply horizontally-

-Distribute

Example (pg.341 #41)

(3X²-2) (X²+4X+3) → 3X²(X²+4X+3) + -2 (X²+4X+3)

= 3X4+12X³+9X² + -2X²-8X-6

= 3X4+12X³+7X²-8x-6

~Special Patterns~~Sum and Difference~

(a+b)(a-b)=a²+b² → (X+7)(X-7)=X²+49 (pg342#53)

~Square of a Binomial~

(a+b)²=a²+2ab+b²→(X+4)²=X²+2(X4)+4²= X²+8X+16(pg.342 #54)

(a-b)²=a²-2ab+b²→(6-X)²= 6²-2(6)(-X)+X²=36+12X+X²=X²+12X+36

~Cube of a Binomial~

(a+b)³=a³+3a²b+3ab²+b³

→(X+4)³=X³+3(X²)(4)+3(X)(4²)+4³= X³+12X²+48x+64

(a-b)³=a³-3a²b+3ab²-b³

→(X-2)³=X³-3(X²)(2)+3(X)(2²)-2³= X³-6X²+12X-8

~Life Application?~

Finding Area:

Ex: Finding Unknown area or volume

Ex:Projectile Motion

The END!I hope you

understand now :)

6.5-POLYNOMIAL DIVISION, FACTORING AND FINDING ZEROS

McKenzie Mahn

POLYNOMIAL LONG DIVISION

In polynomial long division one polynomial, f(x), is being divided by another, g(x) for a quotient, d(x). If a remainder, h(x), is present it is expressed as h(x) g(x)

Algebraically - f(x) = d(x) + h(x) g(x) g(x)

Remainder Theorem- If f(x) is divided by x-k then the remainder will be r=f(k)

EXAMPLEDivide (x² + 7x - 5) by (x – 2)

x – 2) x² + 7x – 5 x²/x=x x² - 2x x(x -2) = x²-2x 9x – 5 (x²+7x-5)-(x² -2x)= 9x-5 9x -18 9x/x =9 and 9(x-2)

=9x-18 13 (9x-5)–(9x-18)= 13 and the

x + 9

Remainder is 13X-2

x+9+13X-2

Answer-

SYNTHETIC DIVISION In synthetic division only the coefficients

of each term are used. The divisor must be in the form (x-k) for

synthetic division to work.

ax²+bx+c divided by x-d X-d=0 = x=d a b c

a (b+d(a)) c+(d(b+d(a))) = ax + (b+d(a)) +

d d(a) d(b+d(a))

c+(d(b+d(a)))X-d

+ +

EXAMPLEDivide (4x² +5x – 4) by (x + 1) x+1=0 x=-1 4 5 -4 + +

4 1 -5 Answer : 4x +1 - 5

-1 -4 -1

X+1

FACTORING WITH SYNTHETIC DIVISION

Synthetic division is used to factor a polynomial f(x) when a factor, (x-k), is given.

coefficients of f(x) k + +

simplify answer “undistribute” or Tic-Tac-Toe

x+k is a factor of f(x), so f(-k)=0 or x=-k

EXAMPLE Factor the polynomial 2X³+11x²+18x+9; f(-3)=0 2x 3

2 11 18 9 x 3x

1 2x-3 -6 -15 -9

2 5 3 0 = (x+3)(2x²+5x+3)

+ + +

3

2x²

5x

2X³+11x²+18x+9= (x+3)(2x+3)(x+1)

FINDING ZEROSTo find the zeros of a polynomial, factor the polynomial completely.

IF - Factored form (x+a)(x-b)(x-c)

THEN - Zeros are –a,b,c

EXAMPLEGiven one zero of the polynomial function, find the other zeros 9x³+10x²-17x-2; -2

9 10 -17 -2

-2 -18 16 2

9 -8 -1 0

+ + + 9x² -9x

x

9x

-1

1 x -1

-8x

9x²-8x-1 =(x+2)(x-1)(9x+1)

Zeros: -2, 1, -1/9

REAL LIFE EXAMPLE

A garden has an area of 10x +5x³+4x²-9. The width of the garden is x+1. What is the length of the garden?

10 5 4 0 -9

10 -5 9 -9 010x³-5x²+9x-9= garden length

4

-1 -10 5 -9 9

+ + + + or undistribute the x

from the first three terms for x(10x²-5x+9)-9

SECTION 6.6FINDING RATIONAL ZEROS

Emily Slade

•  

1. List the possible rational zeros

-12 is the constant term1 is the leading coefficient

2. Find all factors of the coefficient for the numerators1, 2, 3, 4, 6, 12

3. Find all factors of the leading coefficient for the denominator1

4. The possible rational zeros are ±1/1, ± 2/1, ±3/1, ±4/1, ±6/1, ±12/1

Use synthetic division to test the zeros:

Test any of the possible rational zeros

X=1

Because the last number is not 0, 1 is nota zero of f

X= -1

The last number is 0, so -1 is a zero of f

X = -1

f(x)= (x + 1)(x2 + x – 12)

X + 1 because x = -1 when x + 1 = 0

= (x2 + x – 12)

Factor the trinomial:

f(x) = (x + 1)(x2 + x – 12) = (x + 1)(x – 3)(x + 4)

Zeros of f are -1, 3, and -4

x + 1 = 0 x = -1x – 3 = 0 x = 3x + 4 = 0 x = -4

PRACTICE PROBLEMS

Find all real zeros of the function:

1. f(x) = x3 – 8x2 – 23x + 30

= (x + 3)(x – 1)(x – 10)

Zeros of f are -3, 1, and 10

X = -3

x + 3 = 0 x = -3x – 1 = 0 x = 1x – 10 = 0 x = 10

30 is the constant term- factors: 1, 2, 3, 5, 61 is the leading coefficient- factors: 1

Possible rational zeros:±1/1, ±2/1, ±3/1, ±5/1, ±6/1

f(x) =(x + 3)( x2 – 11x + 10)

40 is the constant term- factors: 1, 2, 4, 5, 8, 10, 20, 401 is the leading coefficient- factors: 1

Possible rational zeros: ±1/1, ±2/1, ±4/1, ±5/1, ±8/1, ±10/1, ±20/1, ±40/1

X = -2

f(x) = (x + 2)(x2 – 9x + 20)

= (x + 2)(x – 4)(x – 5)

Zeros of f are -2, 4, and 5

x + 2 = 0 x = -2x – 4 = 0 x = 4x – 5 = 0 x = 5

2. f(x) = x3 – 7x2 + 2x + 40

Possible rational zeros: ±1/1, ±2/1, ±4/1, ±1/2, ±2/2, ±4/2

X = -2

f(x) = (x +2)(2x2 + 0x -2)

= (x + 2)(x +1)(x – 1)

x + 2 = 0 x = -2x + 1 = 0 x = -1x – 1 = 0 x = 1

Zeros of f are -2, -1, 1

3. f(x) = 2x3 + 4x2 – 2x -4

-4 is the constant term- factors: 1, 2, 42 is the leading coefficient- factors: 1, 2

REAL LIFE EXAMPLECreate a model of the pyramid-shaped building at the Louvre Museum using rational zeros

The height of the model will be 2 inches less than the length of each side of the pyramid’s base. What would the dimensions be ?

Volume is V = 1/3Bh

Volume = 25Side of square base = xArea of base = x2

Height = x – 2

25 = 1/3x2(x – 2)

Multiply each side by 375 = x3 – 2x2

Subtract 75 from each side0 = x3 -2x2 - 75

Possible rational zeros: ±1/1, ±3/1, ±5/1, ±15/1, ±25/1, ±75/1

X = 5 is a solution

X2 + 3x + 15 = 0

The base of the model is 5 inches by 5 inches. The height of the mold should be 5 -2 = 3 inches

7.1 Presentation• Project for this section was not

turned in!• Use my lesson: oclick here for PDF version!oclick here for Power Point version!

7.2• Included with Ryan’s 6.1

Presentation!

Chapter 7: Section 3Amanda Giroux

Goal of the Section• In Chapter 6, we learned how to add, subtract,

multiply, and divide polynomial functions. • These operations can be defined for any

functions.

Operations on Functions

• Addition: h(x)= f(x) + g(x)• Subtraction: h(x)= f(x) - g(x)• Multiplication: h(x)=f(x) * g(x)• Division: h(x)=f(x)/g(x)

• The domain of h consists of the x-values that are in the domains of both f and g. Additionally, the domain of a quotient does not include x-values for which g(x)= 0.

Adding Function Example• Let f(x) = 4x1/2 and g(x) = -9x1/2 to find f(x) + g(x)• Step 1:

• Write out the equation• 4x1/2 + (-9x1/2)

• Step 2:• Solve the equation

• f(x) + g(x) = -5x1/2

• Step 3:• Find the domain by making the answer equal to zero and

solve again: (-5x1/2 = 0)• -5√x x≥ 0 all nonnegative real numbers

Subtraction Function Example• Let f(x) = 4x1/2 and g(x) = -9x1/2 to find f(x) - g(x)• Step 1:

• Write out the equation• 4x1/2 - (-9x1/2)

• Step 2:• Solve the equation

• f(x) - g(x) = 13x1/2

• Step 3:• Find the domain by making the answer equal to zero and

solve again: (13x1/2 = 0)

• 13√x x≥ 0 all nonnegative real numbers

Division Function Example• Let f(x) = 6x and g(x) = x3/4 to find f(x)/g(x)• Step 1:

• Write out the equation• 6x / (x3/4)

• Step 2:• Solve the equation

• f(x) / g(x) = 6x1/4

• Step 3:• Find the domain by making the answer equal to zero and

solve again: (6x1/4 = 0)• Since it is an even root and because the denominator

can’t be zero, x has to be any positive real number

Composition of Two Functions• The composition of the function f with the function

is g is: h(x) = f(g(x))

• The domain of h is the set of all x-values such that x is in the domain of g and g(x) is in the domain of f.

Composition of Two Functions Example

• Let f(x) = 2x-7 and g(x) = x2 + 4. What is the value of g(f(3))?

A. -5 B. -3 C. 3 D. 5• Step 1:

• Set up the equation and find g(x)• g(f(3)) g(2(3)-7) g(-1)

• Step 2:• Plug g(-1) into the equation to solve

• (-12+4)= 5

• The answer is D

ALGEBRA 2Chapter 7.4 Review Guide

Inverse Functions

Warm Up

Solve for y:

1) 5y = 10x

2) 3y = 6x + 4

3) 8y = 2x + 1

Inverse Functions

Functions f and g are inverses of each other provided:

f(g(x)) = x and g(f(x))=x

The function g is denoted by f-1 , read as “f inverse”

Inverse Functions

Horizontal Line Test

If no line intersects the graph of a function f more than once, then the inverse of f is itself a function.

Inverse Relations

Sample Problems

1) Find the Inverse Relation:

2) Find the Inverse Function:f(x) = -4x3 + 2

3)Graph the Function of f.f(x) = x2 + 2

X 2 5 7

Y 9 4 1

7.5&7.6By Peter Utz

Warm Up

•1.) Multiply•(x+7) (x+7)•2.) Factor•x2+6x+9•3.) Solve when x=3• (4x-3)

Graphing square root and cube root functions•In 7.4 you learned graphs of rational

functions.•This lesson will teach you how to graph in

the form of y=a (x-h)-k

How to graph

•to graph y= (x-h)-k, first graph y= (x) the shift the graph h units horizontally and k units vertically

Practice

•Graph:•y= (x+5)-2

Practice

•Graph:•y= (x+5)-2First graph (x)

Practice

•Graph:•y= (x+5)-2First graph (x)Then shift the graph 5 units left (inside lies)and 2 units down

How to solve radical equations

•Obtain a polynomial equation•Do so by eliminating radicals (raise each side to the same power)

Practice

Solve (x) -5

Practice

(x) =5First add five to put radical alone

Practice

(x) =5First add five to put radical alone

(x) 3= 53

Then raise both sides to the 3rd power to cancel 3rd root

Practice

(x) =5First add five to put radical alone

(x) 3= 53

Then raise both sides to the 3rd power to cancel 3rd root

x=125 Lastly, solve for each side

Real life

•Graphing square root and cube root functions can be a part of everyday life

•To find how far away the horizon is from your current point, the equation d= (1.5h) h being the height of your viewpoint

•If graphed, the distance of the horizon could be found by simply following the line to your current altitude

Real life…again cuz I picked 2 sections•Many equations used in real world

problem solving include the simple area of a square

•For example, area is equal to the side of the square times the other side of the square

•To find the length of the side, take the square root of the area

9.2

Graphing Simple Rational Functions

A rational function is a function of the form:F(x) = p(x) q(x) Both p(x) and q(x) are polynomials Q(x) cannot be 0

What are Rational Functions?

A graph of this function is called a hyperbola It consists of two asymptotes (vertical and horizontal) and

two symmetrical parts called branches

What Does the Graph of a Rational Function Look Like?

BranchesHorizontal Asymptote

Vertical Asymptote

Hyperbola

There are two different rational function forms: y = + k y = For the y = + k form, the vertical asymptote is x=h and

the horizontal asymptote is y=k For the y = form, the vertical asymptote occurs at the x-

value that makes the denominator zero. The horizontal asymptote is the line y =

Finding the Asymptotes

In order to graph rational functions, you need to plot the horizontal asymptote and vertical asymptote.

Ex.a) y = - 1

b) y = + 5

c) y =

V.A. = -4 H.A. = -1

Graphing the Asymptotes

V.A. = 7 H.A. = 5

V.A. = -2 H.A. =

After plotting the vertical and horizontal asymptotes, you will need to find the branches

To find the branches, it’s best to construct an x/y table and take x points from the left of the asymptote and also to the right of the asymptote and plug them into the equation

Ex. Graph y = +2

Graphing the Branches

x y

1 4

1 3

0 2

3 1

3 1

4 1

The last thing you want to do when graphing simple rational functions is to state the domain and range

The domain of a function is the set of all possible input values (x), which allows the function formula to work so the domain of simple rational functions will be all real numbers except the vertical asymptote

The range of a functions is the set of all possible output values (y), which result from using the function formula so the range of simple rational functions will be all real numbers except the horizontal asymptote

Stating the Domain and Range

a) y = -1 b) y =

Practice Problems

x y x y

You can use the graph of a rational function to solve real-life problems, such as finding the average cost per calendar given the “one-time” charge of the digital images and the unit cost of printing each calendar

You can also use this to solve real-life problems, such as finding the frequency of an approaching ambulance siren

Real Life Example

Algebra 2Section 9.3

Nick Namin

f (x ) = =a m x

m + a m – 1x m – 1 + … + a 1x + a 0

b n x n + b n – 1x

n – 1 + … + b 1x + b 0

p (x )

q (x )

Section 9.3 Overview

Equation

In lesson 9.2, you learned how to graph rational functions of the form

f(x)= p(x) q(x)

Let p(x) and q(x) be polynomials with no common factors other than 1

The graph of the rational function has the following characteristics:

1. The x-intercepts of the graph of f are the real zeros of p(x)2. The vertical asymptotes are at each real zeroes of q(x)3. The horizontal asymptotes:

• If m<n, y=o• If m=n, y=am

bn

• If m>n, no horizontal asymptote

Graph y= 4 x2 + 1

• 4=0 – no x intercepts• x2 + 1= 0 - 1 -1 x2 = -1*If you were to square root -1, the answer would be imaginary*• 1<2, y=0 – horizontal asymptote

X -3 -2 -1 0 1 2 3

Y 2/5 4/5 2 4 2 4/5 2/5

Graph y= 3x2

x2 - 4

3x2 =0 x intercept is 0x2 – 4 = 0 +4 +4 x2 = 4 x=2, -2-vertical asymptotesy=3-horizontal asymptote

X -4 -3 -1 1 3 4

Y 4 5.4 -1 -1 5.4 4

Graph y= x2 -2x- 3 x+4

x2 -2x-3=0(x+1)(x-3) x= -1,3x+4=0 x=-4 vertical asymptote2>0 no horizontal asymptote

X -12 -9 -6 -2 0 2 4 6

Y -20 -19.2 -22.5 2.5 -.075 -.05 .63 2.1

From 1980 to 1995, the total revenue R(in billion of dollars) from parking and automotive service and repair in the United States can be modeled by:

R= 427x2 -6416x+30,432 -.7x3 + 25x2 – 268x +1000

Where x is the number of years since 1980. Graph the model. In what year was the total revenue approximately $75 billion?

Section 9.4

Multiplying and Dividing Rational Expressions

By: Jeffrey Tryon

Simplifying a Rational Expression

x2+12x = x(x+6) = x+6 x2 x . x x

• First factor out numerator and denominator.• Second, divide out factors common to both

the numerator and the denominator.

Multiplying Rational Expressions Involving Polynomials

10x3y . 6x2y2 = 60x5y2

5xy2 10y2 50xy4

= 6 . 10 . x . x4 . y2

5 . 10 . x . y2. y . y = 6x4

5y2

First multiply numerator and denominator

Factor and divide out common factors

Simplified form

Dividing Rational Expressions

x2 + 3x – 40 ÷ x2 + 2x – 48 = x2 + 3x – 40 . x2 + 3x – 18

x2 + 2x – 35 x2 + 3x – 18 x2 + 2x – 35 x2 + 2x – 48

= (x+8)(x-5) . (x+6)(x-3)

(x+7)(x-5) (x-8)(x-6)

= (x+6)(x-3) (x+7) (x-6)

multiply by the reciprocal

Factor then divide out common factors

Simplified form

Real life use!

Rational expressions are used in real life to find a skydivers terminal velocity, which is their constant falling speed.

9.5

Addition, Subtraction, and Complex Fractions

1 GOAL

Working With Rational Expressions

Adding or subtraction rational expressions depends on whether the expressions have like or unlike denominators.If the expressions have like denominator, simply add or subtract their numerators and place the result over the common denominator.

EXAMPLE

Adding and Subtracting with Like Denominators

Perform the indicated operation.

1.2x5

- 8x+1

5= 2x-(8x+1)

5=

2x-8x-1

5=

-6x-1 5

Add numerators and simplify expression

2. 2x

9+

6x+1 2x

=6x+102x

=3x+5

2x

Minus numerators and simplify expression

3. 5+3x

x+1

+-2x-3 x+1

=5+3x-2x-3 x+1

= x+2

x+1

Add numerators and simplify expression

Process to work with unlike denominatorFind the least common denominator of the rational expressions.

Rewrite each expression as an equivalent rational expression using the LCD and proceed as with rational expressions with like denominators.

EXAMPLE

Subtracting With Unlike Denominators

Subtract:

7

6(x-2)-

x+3

6x

7

6(x-2)-

x+3

6x

SOLUTION

= 7x

6x(x-2)-

(x+3)(x-2)6x(x-2)

= 6x(x-2)

7x-(x+3)(x-2)

=7x-(+x-6)

6x(x-2)

= 7x--x+6

6x(x-2)

=-

6x(x-2)

First find the least common denominator of and

7

6(x-2) x+3

6x

It helps to factor each denominator

EXAMPLE

Add With Unlike Denominators

Add:

6x+1

-9+

x

x-3

6x+1

-9+

x

x-3

SOLUTION

= 6x+1

(x+3)(x-3)

+ x(x+3)(x-3)

= (x+3)(x-3)

7x+1

Process to solve complex fractionSimplify the complex fraction

1.Write its numerator and its denominator as single fraction

2.Divide by multiplying by the reciprocal of the denominator

EXAMPLE

Simplifying a Complex Fraction

x+120

14

- 7x+1

Simplify:

x+120

14

- 7x+1

SOLUTION

*x+1x+1

= x+1

20 * 4(x+1

)14

*4(x+1)- 7x+1

*4(x+1)

=20*4

x+1-4*7

= 80

x-27

Simplify , and x+1

20

14

7 x+1

The LCM of them is 4(x+1)

EXAMPLE

Simplifying a Complex Fraction

Simplify:

x

2

- 5

6 - 3x

x

2

- 5

6 - 3x

SOLUTION

=

* 2x

2x

x

2

*2x – 5*2x

6 * 2x- *2x

3x

=- 10x

12x -6

The end

http://prezi.com/5fsdh-onc_gn/section-lesson/

9.6 Prezi – By Meredith