5.4 Prezi – By Anna http://prezi.com/7celq_oomduu/algebra- two-section-54/

5.4 Prezi – By Anna two-section-54/

Chapter 5.5 By Corey Lorraine

Warm Up Question

Vocabulary

Example 1

Example 2

Example 3

Quadratic Formula and The Discriminant
Chapter 5 Section 6 By, 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 = 0 a = 5, b = 6, c = 1 x = (-6 ± √(6^2 - 4×5×1)) / (2×5) x = (-6 ± √( )) / 10 x = (-6 ± √(16)) / 10 x = (-6 ± 4) / 10 x = (-6 – 4) / 10 or x = (-6 + 4) / 10 x = -1 or x = -0.2

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

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

Section: 5.7 Graphing and Solving Quadratic Inequalities
By: Andrew Fratoni

Quadratic Inequalities
Quadratic inequality in two variables Quadratic inequality in one variable

Graphing a Quadratic Inequality Step: 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) + 1 0 < -3 false Shade 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 + 3 Step 2: since the inequality symbol is ≤, make the line solid Step 3: test a point like (0,0) 0 ≤ - (0) ≤ 3 True y ≥ x2 + 2x – 4 Step 1: graph y = x2 + 2x – 4 Step 2: since the inequality symbol is ≥, make the line solid Step 3: test a point like (0,0) 0 ≥ (0)2 + 2(0) – 4 0 ≥ -4 True The shaded region is the area where both parabolas would be shaded and that satisfy both inequalities.

Solving Quadratic Inequalities Algebraically
First, 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 ≥ 0 X2 + 3x – 18 = 0 replace w/ equals sign (x + 6)(x -3) = 0 factor x = -6 x = Set both equal to zero Test x = 0 (0)2 + 3(0) – 18 ≥ 0 -18 ≥ 0 false So, 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 Properties of Exponents
6.1

Properties of Exponents
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

Evaluating Numerical Expressions
(25)3 = ? 215= 32,768 (3/4)2 =? (32/42) = 9/16 (4)3. (4)-6 =? (4) -3= 1/4 3 = 1/64

Student Practice 1.) 52 . 52 2.) (3-3)4 3.) (2/4)3 4.) 50.63 1.) 625
2.) 1/531,441 3.) 1/8 4.) 108

Properties of Rational Exponents
(7.2)

Properties of Rational Exponents
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

1.) n√a . n√b 2.) n√a/ n√b 1.) n√a . b = ? 2.) n(√a/b) = ?
Student Practice 1.) n√a . b = ? 2.) n(√a/b) = ? 1.) n√a . n√b 2.) n√a/ n√b

6.2 evaluating and graphing
polynomial function ALGEBRA 2

a polynomial function is a function of the form
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.

Here is a summary of common types of polynomial function
degree type stand form constant F(x)=a0 1 linear F(x)=a1x+a0 2 Quadratic 3 cubic 4 quartic

Solution 3 0 -8 5 -7 4 f(4)=653 x-value Example 1
use synthetic substitution to evaluate f(x)=3 𝑥 4 −8 𝑥 2 +5x-7 when x=4 Solution coefficients polynomial in standard form 4 f(4)=653 x-value

Example 2 find the recharge time of battery after 100 flashes, if the modeled is where t(in sec) is t= 𝑛 𝑛 n+5.3.the time of a battery to recharge after flashing n times of uses. 100 the recharge time is about 17 seconds

Graph function 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)=x f (x)=-x f (x)→ -∞ As x→ -∞ As x→ +∞

Solution a. 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. -1 -2 -3 X 3 2 1 f(x) -16 32 9 The degree is odd and the leading coefficient is positive so f(x) → -∞ as x →-∞ and f(x) → +∞ as x →+∞

solution b. 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 end 1 8 -15 f(x) -1 -2 -3 x -32 3 2 -147 The degree is even and the leading coefficient is negative so f(x) → -∞ as x →-∞ and f(x) → -∞ as x →+∞

Adding,Subtracting, Multiplying Polynomials J.W.Tang
Algebra II Lesson 6.3 Adding,Subtracting, Multiplying Polynomials J.W.Tang 44

~Adding Polynomials~ To add polynomials Example (pg 341 #25)
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 45

~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__ X³-4X²+3X__ - X³ - X² __ → X³ + X² __ -1 9X³-3X²+3X X³-3X²+3X+1 46

~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 47

-To multiply Vertically-
Example (pg.341 #41) (3X²-2) (X²+4X+3) X²+4X+3 * X²-2 -2x²-8X-6 + 3X412x³+9x²__ ________________ 3X412x³+7X²-8X-6 48

-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 49

~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 50

~Life Application?~ Finding Area: Ex: Finding Unknown area or volume
Ex:Projectile Motion

I hope you understand now :)
The END! I hope you understand now :) 52

6.5- POLYNOMIAL division, Factoring and Finding zeros
McKenzie Mahn 6.5- POLYNOMIAL division, Factoring and Finding zeros

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)

Example Divide (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 (9x-5)–(9x-18)= 13 and the x + 9 Remainder is 13 X-2 Answer- 13 X-2 x+9+

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

Example Divide (4x² +5x – 4) by (x + 1) x+1=0 x= Answer : 4x 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 x 3x 1 2x = (x+3)(2x²+5x+3) 5x 2x² 3 2X³+11x²+18x+9= (x+3)(2x+3)(x+1)

Finding Zeros To 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

Example Given one zero of the polynomial function, find the other zeros 9x³+10x²-17x-2; -2 x -1 -8x 9x 9x² x x -1 Zeros: -2, 1, -1/9 9x²-8x-1 =(x+2)(x-1)(9x+1)

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? x³-5x²+9x-9= garden length 4 or undistribute the x from the first three terms for x(10x²-5x+9)-9

The End

Finding rational zeros
Section 6.6 Finding rational zeros Emily Slade

1. List the possible rational zeros -12 is the constant term
1. List the possible rational zeros -12 is the constant term 1 is the leading coefficient 2. Find all factors of the coefficient for the numerators 1, 2, 3, 4, 6, 12 3. Find all factors of the leading coefficient for the denominator 1 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 not a zero of f X= -1 The last number is 0, so -1 is a zero of f

Zeros of f are -1, 3, and -4 X = -1 = (x2 + x – 12)
f(x)= (x + 1)(x2 + x – 12) X + 1 because x = -1 when x + 1 = 0 Factor the trinomial: f(x) = (x + 1)(x2 + x – 12) = (x + 1)(x – 3)(x + 4) x + 1 = x = -1 x – 3 = x = 3 x + 4 = x = -4 Zeros of f are -1, 3, and -4

Practice problems Find all real zeros of the function:
f(x) = x3 – 8x2 – 23x + 30 30 is the constant term- factors: 1, 2, 3, 5, 6 1 is the leading coefficient- factors: 1 Possible rational zeros: ±1/1, ±2/1, ±3/1, ±5/1, ±6/1 X = -3 f(x) =(x + 3)( x2 – 11x + 10) = (x + 3)(x – 1)(x – 10) x + 3 = x = -3 x – 1 = x = 1 x – 10 = 0 x = 10 Zeros of f are -3, 1, and 10

Zeros of f are -2, 4, and 5 X = -2 2. f(x) = x3 – 7x2 + 2x + 40
40 is the constant term- factors: 1, 2, 4, 5, 8, 10, 20, 40 1 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) x + 2 = x = -2 x – 4 = x = 4 x – 5 = x = 5 Zeros of f are -2, 4, and 5

Zeros of f are -2, -1, 1 3. f(x) = 2x3 + 4x2 – 2x -4
Possible rational zeros: ±1/1, ±2/1, ±4/1, ±1/2, ±2/2, ±4/2 -4 is the constant term- factors: 1, 2, 4 2 is the leading coefficient- factors: 1, 2 X = -2 f(x) = (x +2)(2x2 + 0x -2) = (x + 2)(x +1)(x – 1) x + 2 = x = -2 x + 1 = x = -1 x – 1 = x = 1 Zeros of f are -2, -1, 1

Real life example Create 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 = 25 Side of square base = x Area of base = x2 Height = x – 2 25 = 1/3x2(x – 2) Multiply each side by 3 75 = x3 – 2x2 Subtract 75 from each side 0 = 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: click here for PDF version! click here for Power Point version!

7.2 Included with Ryan’s 6.1 Presentation!

Chapter 7: Section 3 Amanda 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) = x What is the value of g(f(3))? A B C 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

Chapter 7.4 Review Guide Inverse Functions
Algebra 2 Chapter 7.4 Review Guide Inverse Functions

Warm Up Solve for y: 5y = 10x 3y = 6x + 4 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 Find the Inverse Relation: 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&7.6 By Peter Utz

Warm Up 1.) Multiply (x+7) (x+7) 2.) Factor x2+6x+9 3.) Solve when x=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)-2 First graph (x)

Practice Graph: y= (x+5)-2 First 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) =5 First add five to put radical alone

(x) =5 (x) 3= 53 Practice First add five to put radical alone
Then raise both sides to the 3rd power to cancel 3rd root

(x) =5 (x) 3= 53 x=125 Practice First add five to put radical alone
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

Graphing Simple Rational Functions
9.2 Graphing Simple Rational Functions

What are 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 Does the Graph of a Rational Function Look Like?
A graph of this function is called a hyperbola It consists of two asymptotes (vertical and horizontal) and two symmetrical parts called branches Horizontal Asymptote Hyperbola Branches Vertical Asymptote

Finding the Asymptotes
There are two different rational function forms: y = 𝑎 𝑥−ℎ + k y = 𝑎𝑥+𝑏 𝑐𝑥+𝑑 For the y = 𝑎 𝑥−ℎ + k form, the vertical asymptote is x=h and the horizontal asymptote is y=k For the y = 𝑎𝑥+𝑏 𝑐𝑥+𝑑 form, the vertical asymptote occurs at the x- value that makes the denominator zero. The horizontal asymptote is the line y = 𝑎 𝑐

Graphing the Asymptotes
In order to graph rational functions, you need to plot the horizontal asymptote and vertical asymptote. Ex. y = −1 𝑥 y = −5 𝑥 −7 + 5 y = 𝑥+2 2𝑥+4 V.A. = -4 H.A. = -1 V.A. = 7 H.A. = 5 V.A. = -2 H.A. = 1 2

Graphing the Branches 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 = −1 𝑥 −2 +2 x y 1 1 2 4 1 3 2 1 2 3 1 2 1 1 3

Stating the Domain and Range
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

Practice Problems a) y = 2 𝑥 − b) y = 2𝑥+2 2𝑥+1 x y x y

Real Life Example 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

Algebra 2 Section 9.3 Nick Namin

Section 9.3 Overview f (x ) = = p (x )
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 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 ) Equation

The graph of the rational function has the following characteristics:
The x-intercepts of the graph of f are the real zeros of p(x) The vertical asymptotes are at each real zeroes of q(x) The horizontal asymptotes: If m<n, y=o If m=n, y=am bn If m>n, no horizontal asymptote

Graph y= x2 + 1 4=0 – no x intercepts x2 + 1= 0 x2 = -1 *If you were to square root -1, the answer would be imaginary* 1<2, y=0 – horizontal asymptote X -3 -2 -1 1 2 3 Y 2/5 4/5 4

Graph y= 3x2 x2 - 4 3x2 =0 x intercept is 0 x2 – 4 = 0 x2 = 4 x=2, -2-vertical asymptotes y=3-horizontal asymptote X -4 -3 -1 1 3 4 Y 5.4

Graph y= x2 -2x- 3 x+4 x2 -2x-3=0 (x+1)(x-3) x= -1,3 x+4=0 x=-4 vertical asymptote 2>0 no horizontal asymptote X -12 -9 -6 -2 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= 427x x+30,432 -.7x3 + 25x2 – 268x +1000 Where x is the number of years since Graph the model. In what year was the total revenue approximately $75 billion?

Multiplying and Dividing Rational Expressions
Section 9.4 Multiplying and Dividing Rational Expressions By: Jeffrey Tryon

Simplifying a Rational Expression
x2+12x = x(x+6) = x+6 x 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 = x . x4 . y x . y2. y . y = 6x4 5y2 First multiply numerator and denominator Factor and divide out common factors Simplified form

Dividing Rational Expressions
multiply by the reciprocal x2 + 3x – 40 ÷ x2 + 2x – 48 = x2 + 3x – x2 + 3x – 18 x2 + 2x – x2 + 3x – x2 + 2x – 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) 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 Perform the indicated operation.
Adding and Subtracting with Like Denominators Perform the indicated operation. 2x 5 - 8x+1 = 2x-(8x+1) 2x-8x-1 -6x-1 Minus numerators and simplify expression 1. 2x 9 + 6x+1 = 6x+10 3x+5 2. Add numerators and simplify expression 5+3x x+1 + -2x-3 = 5+3x-2x-3 x+2 3. Add numerators and simplify expression

Process to work with unlike denominator
Find 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 6(x-2) x+3 6x Subtract:
7 6(x-2) - x+3 6x Subtract: SOLUTION 7 6(x-2) - x+3 6x 7x (x+3)(x-2) = - 6x(x-2) 6x(x-2) First find the least common denominator of and 7 x+3 6x 7x-(x+3)(x-2) = 6(x-2) 6x(x-2) 7x-( 𝑥 2 +x-6) = It helps to factor each denominator 6x(x-2) 7x- 𝑥 2 -x+6 = 6x(x-2) = - 𝑥 2 +6x+6 6x(x-2)

EXAMPLE Add With Unlike Denominators 6x+1 + Add: SOLUTION 6x+1 + 6x+1
𝑥 2 -9 + x x-3 Add: SOLUTION 6x+1 𝑥 2 -9 + x x-3 6x+1 x = + (x+3)(x-3) (x+3)(x-3) 7x+1 = (x+3)(x-3)

Process to solve complex fraction
Simplify the complex fraction Write its numerator and its denominator as single fraction Divide by multiplying by the reciprocal of the denominator

EXAMPLE Simplifying a Complex Fraction 1 - 7 Simplify: 1 7 SOLUTION
20 1 4 - 7 Simplify: 1 4 7 SOLUTION Simplify , and x+1 20 x+1 20 x+1 x+1 * 1 4 7 x+1 - The LCM of them is 4(x+1) x+1 x+1 20 * 4(x+1) = 1 4 7 *4(x+1)- *4(x+1) x+1 20*4 = x+1-4*7 80 = x-27

EXAMPLE Simplifying a Complex Fraction Simplify: x - 5 3 6 - SOLUTION
2 - 5 6 - 3 Simplify: SOLUTION x 2 - 5 2x * 3 2x 6 - x x 2 *2x – 5*2x = 3 x 6 * 2x *2x 𝑥 x = 12x -6

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