a quadratic function f is a function of the form f(x) = ax 2 + bx + c where a, b and c are real...

• A quadratic function f is a function of the form

f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a.

Jeff White

5.1

Graphing A Quadratic Function

Vertex is (2,-2)Then draw the axis of symmetry which is x=2Then plot two points on one side of the axis of symmetry. Use symmetry to plot two more points.

Jeff White

Graphing A Quadratic Function In Vertex Form

Vertex Form

First plot the vertex (H,K) = (-3,4)Then draw the axis of symmetry X=-3 and plot two points on one side of it.Use symmetry to complete the graph.

Jeff White

Graphing A Quadratic Function In Intercept Form

Intercept Form

X-intercept occur at (-2,0) and (4,0)Axis of symmetry is 1X-coordinate of the vertex is x=1.The y-coordinate of the vertex is

Jeff White

5.4 Complex Numbers• Imaginary unit is called i• i= √ (-1)• r is a positive real number

√ (–r)= i √ (r)

Complex number written in standard form is a+bi a & b real numbers a real part of complex number b imaginary part of complex number

• b ≠0 a+bi is imaginary

• a=0, b≠0 a=bi is pure imaginary number

• z=a+bi is complex number

Sample Problems• 1. Solve

• 3. Write the (2+3i)+(7+i) as a complex number in standard form.

• 5. Divide

• 2. Write (8+5i)-(1+2i) as a complex number in standard form.

• 4. Multiply i(3+i).

6. Find the absolute value of 3-4i

i1

8

Helpful Hints• 1. Take the square root of x squared and -4.• 2. Distribute the minus sign to 1 and 2i. Combine like terms.• 3. Distribute the plus sign to 7 and i. Combine like terms.• 4. Distribute the i to 3 and i.• 5. Divide 8 by 1 and 8 by i.

• 6. Consult the formula on the first page: a=3, b=-4

Answers

• 1. • 2.

• 3. • 4.

• 5. • 6.

Chapter 5.5: Completing the SquareGoal 1: Solving Quadratic Equations by Completing the SquareCompleting the square is a process that allows you to write an expression of the form

x2 + bx as the square of a binomial. To complete the square for x2 + bx, you need to add (b/2) 2. The following is a rule for completing the square: X2 + bx +(b/2)2 = (x+[b/2])2

Example 1: Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.

x2 + 18x + c Write the equation out

b=18 Use the formula to find b

c = (b/2)2 = (18/2)2 = 92 = 81 Find the value of c that makes the expression a perfect square trinomial

x2 + 18x + 81 Substitute the C value in the expression.

(x+9)2 Factor to get your answer

Matt

Chapter 5.5: Completing the Square

Example 2: Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square.X2 + 2x = 9 Write out original equationX2 + 2x + 1 = 10 Add (2/2)2 = 12 = 1 to each side(x+1)2 = 10 Write the left side as a binomial squaredX + 1 = √10 Take the square roots of each sideX = -1 + √10 Solve for x

Example 3: Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square.

6x2 +84x +300 = 0 Write the original equationX2 +14x +50 = 0 Divide both sides by the coefficient of x2

X2 + 14x = -50 Write the left side in the form of x2 + bxX2 + 14x + 49 = -1 Add (14/2)2 = 72 = 49 to each side(X + 7)2 = -1 Write left side as a binomial squaredX + 7 = √-1 Take the square roots of each sideX = -7 ± √-1 Solve for xX = -7 ± i Write in terms of the imaginary unit i

Chapter 5.5: Completing the Square

Goal 2: Writing Quadratic Functions in Vertex Form: Given a quadratic function in standard form, y = ax2 + bx + c, you can use completing the square to write the function in vertex form, y = a(x – h)2 + k.

Example 4: Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex.

Y = x2 – 6x + 11 Write out the original functionY + 9 = (x2 – 6x + 9) + 11 Complete the square of x2 – 6;

add (-6/2)2 = -32 = 9Y + 9 = (x – 3)2 + 11 Write x2 – 6x + 9 as a binomial squared Y = (x – 3)2 +2 Solve for y

Vertex = (3,2)

Chapter 5.5: Completing the SquarePractice Problems for Completing the Square: Find the value of c that makes the expression a

perfect square trinomial. Then write the expression as the square of a binomial.X2 – 44x + cAnswer: C = 484; (x – 22)2

Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square.

X2 + 20x + 104 = 0Answer: -10 + 2i, -10 – 2i

Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square.

2x2 – 12x = -14Answer: 3 ± √2

Practice Problems for Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex.

Y = x2 – 3x – 2Answer: y = (x – [3/2])2 – (17/4)Vertex: ([3/2], [-17/4])

Algebra II Section 5.6

A presentation by:Elise Couillard; Block 5

June 9, 2010

Solving Equations With The Quadratic Formula

• By completing the square once for the general equation , you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula.

The Quadratic Formula

• The Quadratic Formula:*Let a, b, and c be real numbers such that a does not equal 0. The solutions of the quadratic

equation are:

Solving a Quadratic Equation With 2 Real Solutions

• Solve

The solutions are x=1.35 and x=-1.85

Number and Type of Solutions of a Quadratic Equation

• Consider the quadratic equation*If >0, then the equation has two real solutions.*If =0, then the equation has one real solution.*If <0, then the equation has two imaginary solutions.

THE END!Sources: Algebra II Textbook

Lesson 5.7 Jessica Semmelrock

Quadratic Inequality in 2 Variables

Quadratic Inequality in 1 Variable

y < ax2 + bx +c

y > ax2 + bx +c

y < ax2 + bx +c

y < ax2 + bx +c

ax2 + bx +c < 0

ax2 + bx +c > 0

ax2 + bx +c < 0

ax2 + bx +c > 0

1. Graph y < x2 + 8x + 16

Example 1

2. Test the point (0,0)y < x2 + 8x + 160 < 02 + 8(0)+ 160 < 163. Shade outside region because 0 < 16


Graph the system of quadratic inequalities

y > x2

y < x2 + 3

Example 2:

Only difference to this problem is when the

shading overlaps that is your answer.


Solve x2 + x -2 < 0 by graphingExample 3:

Step 1: x2 + x -2 = 0• Find graph intercepts by replacing 0 for x.

Step 2: x = -1 + 1 – 4(1)(-2)

2(1)•Use quadratic formula to solve for x.

Answer: x ≤ –1.37 or x ≥ .37


Solve x2 + 3 -18 > 0 algebraicallyExample 4:

x2 + 3 -18 > 0x2 + 3 -18 = 0(x-3) (x+6) = 0

Answer x = 3 or x = -6

Test an x-value in each interval to see if it satisfies the inequality.

Test these points with the arrows.

Travis Deskus

6.1 / 7.2

Using Properties of Exponents

-

Evaluating Numerical Expressions• 1. Product of like bases: Example: x5 x3 = x5+3 = x8• To multiply powers with the same base, add the exponents and keep the

common base.• =32• 2.Evaluating numerical expresions a.

Simplifying Algebraic Expressions

• a.• b.• c.• Scientific Notation

7.2 Properties of Rational Exponents

• Example 1. • • If m= for some integer n greater than 1, the third and sixth properties can be

written using radical notation as following:

– product property– Quotient property

– a.

– b. – Using Properties of Radicals

Writing Radicals in Simplest Form• a. steps- factor out perfect cube, product property, simplify• Adding and Subtracting Roots and Radicles

• Book Example• The properties of rational exponent and radicals can also be applied to expressions involving

variables. Because a variable can be positive, negative or zero, sometimes absolute value is needed when simplifying a variable expression.

• =x when n is odd = when n is even • Simplifying expressions involving varribles

Section 6.2

IDENTIFYING POLYNOMIAL FUNCTIONS

• A polynomial is a function if it’s in standard form and the exponent is a whole number

• ex: f(x)= 3

• If the polynomial has an exponent that is not a whole number it’s not a function

• ex: f(x)= 3x1/2 – 2x2 +5

Using Synthetic Substitution

• Write the polynomial in standard form• Insert terms with coefficients of 0 for missing terms• Then write the coefficients of f(x) in a row • Bring down the leading coefficients and multiply them by 1• Write results in the next column and bring down your results • Continue until you reach the end of the row • EX:

Graphing Polynomials Functions• Begin by making a table of values, including positive, negative, and zero

values for x• Plot the points and connect them with a smooth curve. Then check the

behavior

• The Degree is odd and the leading coefficients is positive, so • f(x)→ + -• As x → - • f(x) → + • As x → +

X -3 -2 -1 0 1 2 3

f(x) -218 -18 6 4 6 42 262

Examples of Graphs

Work • Graph the polynomial function • 1.) f(x)= x4+3• 2.) g(x)= x3-5• 3.) h(x)= 2+ x2-x4

• Synthetic substitution • 1.) f(x)=x3+ 5x2+4x+6, x=2• 2.) f(x)= x3-x5 +3, x=-1• 3.) f(x)= 5x3-4x2-2, x=0• Functions yes or no• 1.) f(x)= x4+3• 2.) f(x)=5x3/4-5x2+3

Adding, Subracting, and

Multiplying Polynomials

6.3

By: Robert Johnson

How to Solve

●To add or subtract polynomials, add or subtract the coefficients of LIKE terms. You can do this by using a vertical or horizontal format

●To Multiply two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Then combine LIKE terms

Adding Polynomials

Add 2x3-5x2+3x-9 and x3+6x2+11 in vertical format

Add 3x3+2x2-x-7 and x3-10x2+8 in horizontal format

Subtracting Polynomials

Subtract 3x3+2x2-x+7 from 8x3-x2-5x+2 in vertical format

Subract 8x3-3x2-2x+9 from2x3+6x2-x+1 in horizontal format

Multiplying Polynomials

Multiply -2y2+3y-6 and y-2 in vertical format

Multiply -x2+2x+4 and x-3 in horizontal format

Special Product Patterns● Sum and difference● (a+b)(a-b)= a2 - b2

● Square of a Binomial● (a+b)2 = a2 + 2ab + b2

● (a-b)2 = a2 - 2ab + b2

● Cube of a Binomial

● (a+b)3 = a3 + 3a2b + 3ab2 + b3

● (a-b)3 = a3 - 3a2b + 3ab2 - b3

Using Special Product Patterns

(x + 2)(3x2 - x – 5)

(a – 5)(a + 2)(a + 6)

(xy - 4)3

Anjy Grasso

Anjy Grasso

Put the equation into Standard formIn other words, make it equal zero.

x² + 8x = −12

x² + 8x = −12

+12 +12

x² + 8x + 12 = 0

Now that you have the equation you're solving, find the first factor

x² + 8x + 12 = 0

x² + 8x + 12 = 0( ) ( ) = 0the only factors of x2 are x * x, you

now have the first factors.

x² + 8x + 12 = 0(x ) (x ) = 0

Anjy Grasso

Now that we have the first factors, the x2 goes away, and we're left with this:

8x + 12 = 0(x ) (x ) = 0

Now we find the factors of 12.

1 and 12 wont work,neither will 4 and 3,so lets use 2 and 6

8x + 12 = 0(x ) (x ) = 0

(x 6) (x 2) = 0

Since the entire equation was positive,Both of these should be positive too.

(x + 6) (x + 2) = 0

Anjy Grasso

So how exactly do we know this is right? Lets use FOIL (first, outer, inner, last) multiplication to test it out.

(x + 6) (x + 2) = 0

F - First

O - Outer

I - Inner

L - Last

x * x = x2

x * 2 = + 2x

6 * x = + 6x

6 * 2 = + 12

x2 + 2x + 6x + 12Now Simplify…

x2 + 8x + 12

There's Your original equation!

Polynomial Long Division – when dividing a polynomial f(x) by a divisor d(x), you get a quotient polynomial -q(x) and a remainder polynomial r(x) f(x) = q(x) + r(x)

d(x) d(x)

Remainder Theorem – If polynomial f(x) is divided by x-k, then the remainder is r = f(x)

Synthetic Division – only use the Coefficients of the polynomial and the x – k must be in the form of a divisor.

Factor Theorem – A polynomial f(x) has a factor x-k if and only if f(k) = 0

Rachael SKinner

Long Division

f(x) = 3x² + 4x -3 by x² - 3x + 5

* Don't forget to add in exponents if needed exponents must go in numerical order

3x⁴ - 5x³ 0x² + 4x – 6 x² - 3x + 5

3x² + 4x -3

3x⁴- 9x³+ 15x²-

* Remember to subtract – which means the signs will change

4x² - 15x² + 4x4x² - 12x² - 20x-

-3x² - 16x-6-3x² + 9x -15

25x+9

3x² + 4x -325x+9

x² - 3x + 5Remainder

Rachael SKinner

Synthetic Division

f(x) = 2x³ +x² - 8x +5 by x + 3

-3

x + 3 = 0x – 3 = - 3x = - 3

2 1 -8 5

-6 15 -21

2 -5 7 -16

*Use only coefficients

2x³ +x² - 8x +5 -16 x+3

Remainder

Factoring Completely

f(x) = 3x³ - 4x² - 28x – 16 x +2 is a factor x = 2

23 -4 -28 -16

-6 20 16

3 -10 -8 0 no remainder

f(x) (2 + x) (3x² - 10x -8)

Factor

f(x) (x+2) (3x+2) (x – 4)

(x+2) = -2 (3x+2) = -2/3(x – 4) = 4

Solve

finding the 0's of f(x)

Rachael SKinner

1.) Divide 2x⁴ + 3x³ + 5x – 1 by x² – 2x +2 Using Long Division

2.) Divide x³ + 2x² - 6x – 9 by x-2Using Synthetic Division

3.) Factor Completely given that x-4 is a factorf(x) = x³ + 6x² + 5x +12

Rachael SKinner

1.) 2x² + 7x +10 + 11x-21 x² -2x +2

2.) x² + 4x + 2 + -5 x- 2

3.) f(x) = (x-3) (x-4) (x+1)

Rachael SKinner

How to find the rational zeros of a polynomial function

Lesson 6.6

Nate

Rational Zero Theorem

• If f(x)=anxn+…..+a1x+a0 has integer coefficients, then every rational zero of f has the form: P/Q =

factor of constant term a0 / factor of leading coefficient an

Nate

Steps to Solve

• List all possible rational zeros, both positive and negative.

• Change the number that is outside until one of these numbers makes your answer zero.

• When your answer is zero, insert the numbers you got in order to find the value of x.

Nate

Example

Solve: f(x)=3x3-4x2-17x+6

Nate

Answer

3 -4 -17 6 Synthetic Division-2 -6 20 -6

3 -10 3 0 x= 1,2,3,6 (all #’s are +or -) / 1,3 (all #’s are + or -)

f(x)=(x+2)(3x2-10x+3) Factor the trinomialf(x)=(x+2)(3x-1)(x-3) and use the factor

x=-2,3,1/3 theorem

Nate

7.1 nth roots

• n is an integer greater than 1 and a is a real number:

• n is odd, a has one nth root: =a• n is even and a>0 a has 2 nth roots: = a• n is even and a=0 a has 1 nth root: =0 =0• n is even and a<0 a has 0 nth roots

Loren

7.1 nth roots

• a is the nth root of a and m is a positive integer:

• a =(a ) m =( ) m

• a = = m = m , a 0

Loren

Practice Problems

• Rewrite using rational exponent notation.

• Fine the indicated real nth root(s) of an=4, a=0

• Evaluate the expression

Loren

Section 7.3

Power Functions and Function Operations

Andy

1 1

2 2( ) 4 ( ) 9 . .Let f x x and g x x Find the following

.) ( ) ( )a f x g x

= 4x1/2 + (–9x1/2)

= [4 + (–9)]x1/2

= –5x1/2

1. Combine the coefficients

2. Simplify

.) ( ) ( )b f x g x

= 4x1/2 – (–9x1/2)

1. Combine the coefficients= [4 – (–9)]x1/2

2. Simplify

= 13x1/2

Examples

Andy

Practice Problems

f (x) + g(x)1.

2. f (x) – g(x)

2 2

3 3( ) 2 ( ) 4 . .Let f x x and g x x Find the following

Andy

Examples3

4( .( ) 6 ) .Let f x x and g x x Find the following

a. f (x) g(x)b.

f (x)

g(x)

= (6x)(x3/4)

= 6x(1 + 3/4)

= 6x7/4

= 6x

x3/4

= 6x(1 – 3/4)

= 6x1/4

Combine powers by adding

Simplify

Combine powers by subtraction

Simplify

Plug in equation Plug in equation

Andy

Practice Problems

f (x) g(x)4.

f (x)5.

g(x)

1

5( .( ) 4 ) .Let f x x and g x x Find the following

Andy

Examples1( ) 4 ( ) 5 2. .Let f x x and g x x Find the following

a. f(g(x))

= f(5x – 2)

= 4(5x – 2)–1

b. g(f(x))

= g(4x–1)

= 5(4x–1) – 2

= –20x–1 – 2

20

x – 2=

Plug g(x) into the equation

Plug in the coefficient and power of f(x)

Distribute the 5

Simplify

Plug g(x) into the equation

Plug in equation Plug in equation

Andy

Examples1( ) 4 ( ) 5 2. .Let f x x and g x x Find the following

c. f(f(x))

= f(4x–1)

= 4(4x–1)–1

= 4(4–1 x)

= 40xOR

= x

Plug f(x) into the equation

Distribute the exponent

Distribute the 4

Plug in equation

Andy

Practice Problems

a. f(g(x))

b. g(f(x))

c. f(f(x))

2( ) 2 6 ( ) 3 . .Let f x x and g x x Find the following

Andy

7.4

Ileishka Ortiz

Inverse Functions

Inverse relations – maps the output values back to their original input values

•Domain – of inverse relation is the range of the original relation.•Range – of inverse relation is the domain of the original relation.•Inverse of linear functions: reflect original relation over y=x to obtain the inverse relation.

•Which means switch the roles of x & y and solve for y (if possible).

Concepts

7.4

Ileishka Ortiz

Original relation

Inverse relation

RANGE

DOMAIN DOMAIN

RANGE

Ileishka Ortiz

1.) Find an equation for the inverse of the relation y = 2 x – 4

SOLUTION

y = 2 x – 4

x = 2 y – 4

x + 4 = 2 y

x + 2 = y1

2

Write original relation.

Switch x and y

Add 4 to each side.

Divide each side by 2.

7.4

Ileishka Ortiz

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.”

Given any function, you can always find its inverse relation by switching x and y. For a linear function f (x ) = mx + b where m 0, the inverse is itself a linear function.

7.4

Ileishka Ortiz

1.) Verify that f (x) = 2 x – 4 and g (x) = 1/2 x + 2 are inverses.

SOLUTION

Show that f (g (x)) = x and g (f (x)) = x.

g (f (x)) = g (2x – 4)

= (2x – 4) + 2

= x – 2 + 2

= x

12

f (g (x)) = f 1/2 x + 2 ( )

= 2 1/2 x + 2 – 4

= x + 4 – 4

= x

( )

7.4 PRACTICE

Ileishka Ortiz

“If Da Gawd Can Do It Perfect, You

Can Do It Almost Perfect Too” - Mrs.

Delaney

7.5 + 7.6 With Da Gawd

• By: Jon Reyyashi

Its Easy

To graph y=a√x-h + k or y=a^3√x-h + ky=a√x 0r y=a^3√x

H + KH means shift the graph left our right and K means move the point either up or down.

7.5

LET US START WITH 7.5 SHALL WE1. √X+1-3

SO LETS SAY IT IS √X-(-1)+(-3). H=-1 AND K= -3’TO OBTAIN THE GRAPH OF Y=√X+1-3, SHIFT THE GRAPH OF Y=√X LEFT 1 UNIT AND DOWN 3 UNITS

GRAPHING IT

EXAMPLE - Y=-3√X-2 + 1DRAW THE GRAPH OF Y=-3√X

DASHED LINE IN GRAPHIT BEGINS AT THE ORIGIN AND PASSES THROUGH (1, -3)YOU THEN SHIFT THE GRAPH 2 UNITS UP AND UP 1 UNIT.THE GRAPH STARTS (2,1) AND PASSES THROUGH POINT (3, -2)

Domain and Range

State the domain and range of the functionsFrom the graph of y=-3√x-2 + 1 the domain is x≥2 and the range is y≤1

Another Example

ANOTHER EXAMPLE 3^3√X+2 - 11. Sketch the graph of y=3^3√x+2 - 1

This means h=-2 and k=-1 Sketch the graph of y=3^3√x

It passes the origin and points (-1, -3) and (1, 3)You then shift the graph to the left two and down oneYou repeat these steps for each point. Example - (0,0) becomes (-2, -1) because you basically move the point over -2 and down one so it becomes (-2, -1)

Domain and Range

Domain and range are both all real numbers

7.6Powers Property of Equality: If a=B then A^n = B^n

This means that you can raise each side of an equation to the same powerAn extraneous solution is a trial solution that does not satisfy the original equation

Example

√x + 5 = 9Isolate the radical by subtracting 5 by each side

(√x)^2 = 4^2Simplify

x = 16

Radical Exponents

Example3x^3/4 = 192

Isolate the power by dividing each side by 3(x^3/4) = 64

Raise each side by 4/3 power, and cancel the original 4/3 power by performing its reciprocalx = (64^1/3)4

Apply Properties of rootsx = 4^ 4 = 256

Simplify (The solution is 256) Check answer by substituting.

With Only One Radical3√8x+3 - 5 = -2

Isolate the radical by adding 5 to each side3√8x+3 = 3^3

Cube each side8x +3 = 27

Subtract each side of by 38x=24

Simplifyx=3

Two Radicals

3√2x+4 = 2^3√3-xCube each side

(3√2x+4)^3 = (2^3√3-x)^3Simplify

2x + 4 = 8(3-x)Distribute

2x + 4 = 24 - 8x ( Add 8 x to each side10x + 4 = 24x=2

Follow These Rules and

You will be GAWDLY like Me and get a 100 on the Final. Thank You :)

Section 9.3

Y = p (x) / q (x) = amxm+am-1xm-1+…+ax+a0

bnxn+bn-1xn-1+…+bx+b0

Sydney

Section 9.3 - Concepts• P (x) and q (x) are polynomials with no common factors other

than 1.• The graph of the function p (x) /q (x) has the following

characteristics-the x intercepts are the real zeros of p (x) *you set the polynomial p (x) equal to zero and solve*-there is a vertical asymptote at each real zero of q (x) *you set the polynomial q (x) equal to zero and solve*-at most there is one horizontal asymptote

Sydney

Section 9.3 – Concepts cont.

• If m < n, the line y=0 is the horizontal asymptote

• If m = n, the line y = Am / Bn is the horizontal asymptote

• If m > n, the graph has no horizontal asymptote.

Sydney

Section 9.3 – Example 1 (m<n)

• Graph:y = 4

x2 + 1

• Answer:P (x) has no real zeros, so no x-

interceptsThe denominator has no real

zeros, so no vertical asymptote

M<n, so the horizontal asymptote is y = 0

Sydney

Section 9.3 – Example 2 (m=n)

• Graph: y = 3x2

x2-4

• Answer:The numerator’s only zero is

zero. X-intercept is (0,0)Vertical asymptote’s at 2 and -

2M=N so horizontal asymptote

is at 3

Sydney

Section 9.3 – Example 3 (M>N)

• Graph:y = x2-2x-3

x+4

• Answer: X-intercepts are 3 and -1Vertical asymptote is -4M>n, so no horizontal

asymptote

Sydney

9.4

Multiplying and Dividing Rational Expressions

Brett Robinson

Brett Robinson

MultiplyingFactor Numerator and Denominator

Divide out common factors before multiplying if possible

Simplified Form

Brett Robinson

DividingMultiply by the reciprocal

Factor

Divide out common Factors

Simplified Form

Brett Robinson

Section 9.5Addition, Subtraction, and Complex

FractionsAdding and Subtracting with Like Denominators:Solve the following problems:

Step I: Add numerators and simplify the expressions

4

3x

5

3x

4 5

3x

9

3x

3

x

QuickTime™ and a decompressor

are needed to see this picture.

2x

x 3

4

x 3

2x 4

x 3

Step I: Subtract numerators

Jorge Verde

Adding with Unlike Denominators



Solution:

Step I: First find the LCD of and

The LCD is

Step 2: Use this to rewrite each expression.

5

6x2

x

4x2 12x

5

6x2

x

4x(x 3)

5[2(x 3)]

6x2[2(x 3)]

x(3x)

4x(x 3)(3x)

10x 30

12x2(x 3)

3x2

12x2(x 3)

3x2 10x 30

12x2(x 3)

5

6 x 2

x

4 x 2 12 x

12x 2 (x 3)

Jorge Verde

Subtracting with Unlike Denominators



Solution

x 1

x 2 4x 4

2

x 2 4

x 1

(x 2)2 2

(x 2)(x 2)

(x 1)(x 2)

(x 2)2(x 2)

2(x 2)

(x 2)(x 2)(x 2)

x 2 x 2 (2x 4)(x 2)2(x 2)

x 2 3x 6

(x 2)2(x 2)

Jorge Verde

Simplifying a Complex Fraction



Solution

Step I: Add fractions in denominator.

Step 2: Multiply by reciprocal.

Step 3: Divide out common factor.

Step 4: Write in simplified form.









Jorge Verde

Section 9.6

Solving Rational Equations

Morgan Hillhouse

Different ways to Solve Rational Expressions

Cross Multiplying• Set up your equation.• Multiply together the top

expression on the left with the bottom on the right and the top expression on the right with the bottom of the left.

• Set the result equal to zero.• Solve for each part.

Multiple Rational Expressions• Set up your equation.• Multiply each side if the

equation by the LCD of both terms.

• Simplify each side.• Set the simplified form to

zero and solve for each result.

• Check for extraneous solutions.

Morgan Hillhouse

Cross Multiplying

X 3 ― = ―X-4 4

4X=3X-12 -3X

4X=-12

4

X=-3

Morgan Hillhouse

Continued Problem #2

X+6 7―― = ― X 8

8X+48=7X -48 -7X

X=-48

Morgan Hillhouse

Multiple Rational Expressions

CONTINUED ON NEXT SLIDE

2X 4X 17X+4―― - ―― = ――――X-2 3X+2 3X₂-4X-4

(3X+2)(X-2)

(3X+2) (X-2)

2X 4X 17X+4―― - ―― = ――――X-2 3X+2 (3X+2)(X-2)

Morgan Hillhouse

Continued

2X(3X+2)-(4X-1)(X-2)=17X+4

6X₂+4X-(4X₂-9X+2)=17X+4

6X₂+4X-4X₂+9X-2=17+4

2X₂-4X-6=0

-17-4

2(X₂-2X-3)=02(X-3)(X+1)=0X-3=0 X+1=0

X=3 or -1Morgan Hillhouse

a quadratic function f is a function of the form f(x) = ax 2 + bx + c where a, b and c are real...

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