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5.1
• 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
Graphing A Quadratic Function
Vertex is (2,-2)
Then draw the axis of symmetry which is x=2
Then 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 1
X-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
8
1 i
• 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
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 Square
Goal 1: Solving Quadratic Equations by Completing the Square
Completing 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 equation
X2 + 2x + 1 = 10
Add (2/2)2 = 12 = 1 to each side
(x+1)2 = 10
Write the left side as a binomial squared
X + 1 = √10
Take the square roots of each side
X = -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 equation
X2 +14x +50 = 0
Divide both sides by the coefficient of x2
X2 + 14x = -50
Write the left side in the form of x2 + bx
X2 + 14x + 49 = -1
Add (14/2)2 = 72 = 49 to each side
(X + 7)2 = -1
Write left side as a binomial squared
X + 7 = √-1
Take the square roots of each side
X = -7 ± √-1
Solve for x
X = -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 function
Y + 9 = (x2 – 6x + 9) + 11
Complete the square of x2 – 6;
add (-6/2)2 = -32 = 9
Y + 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 Square
Practice 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 + c
Answer: 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 = 0
Answer: -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 = -14
Answer: 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 – 2
Answer: 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
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.
Sources: Algebra II Textbook
THE END!
Quadratic Inequality
in 2 Variables
y < ax2 + bx
+c
y > ax2 + bx +c
y < ax2 + bx +c
y < ax2 + bx +c
Quadratic Inequality
in 1 Variable
ax2 + bx +c < 0
ax2 + bx +c > 0
Example 1
1. Graph
y < x2 + 8x + 16
2. Test the point
(0,0)
y < x2 + 8x + 16
0 < 02 + 8(0)+ 16
0 < 16
3. Shade outside
region because 0 <
Lesson 5.7 Jessica Semmelrock
16
ax2 + bx +c < 0
ax2 + bx +c > 0
Example 2:
Graph the system of quadratic inequalities
y > x2
y < x2 + 3
Only difference to this
problem is when the
shading overlaps that
is your answer.
Lesson 5.7 Jessica Semmelrock
Example 3:Solve x2 + x -2 < 0 by graphing
Step 1: x2 + x -2 = 0
• Find graph intercepts
by replacing 0 for x.
Step 2: x = -1 +
2)
1 – 4(1)(-
2(1)
•Use quadratic formula to solve
for x.
Answer: x ≤ –1.37 or x ≥
.37
Lesson 5.7 Jessica Semmelrock
Example 4:Solve x2 + 3 -18 > 0 algebraically
x2 + 3 -18 > 0
x2 + 3 -18 = 0
(x-3) (x+6) = 0
Test an x-value in each interval to see if it
satisfies the inequality.
Test these points with the arrows.
Answer x = 3 or x = -6
Lesson 5.7 Jessica Semmelrock
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
X
•
•
•
•
•
-3
-2
-1
0
1
2
3
f(x)
-218 is odd
-18 and 6the leading
4
6
42 is positive,
262
The Degree
coefficients
so
f(x)→ + As x → f(x) → +
As x → +
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
x² + 8x + 12 = 0
x² + 8x = −12
Now that you have the equation you're
solving, find the first factor
Put the equation into Standard form
In other words, make it equal zero.
x² + 8x + 12 = 0
(
) (
)=0
x² + 8x = −12
+12 +12
the only factors of x2 are x * x, you
now have the first factors.
x² + 8x + 12 = 0
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
x * x = x2
O - Outer
x * 2 = + 2x
I - Inner
6 * x = + 6x
Now Simplify…
L - Last
6 * 2 = + 12
x2 + 8x + 12
x2 + 2x + 6x + 12
There's Your original equation!
Anjy Grasso
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
3x² + 4x -3
x² - 3x + 5 3x⁴ - 5x³ 0x² + 4x – 6
-3x⁴- 9x³+ 15x²
4x² - 15x² + 4x
-4x² - 12x² - 20x
-3x² - 16x-6
-3x² + 9x -15
25x+9
3x² + 4x -3
25x+9
x² - 3x + 5
Remainder
Rachael SKinner
* Don't forget to add
in exponents if needed
exponents must go in
numerical order
* Remember to
subtract – which
means the signs will
change
Synthetic Division
Factoring Completely
f(x) = 2x³ +x² - 8x +5 by x + 3
x+3=0
x–3=-3
x=-3
f(x) = 3x³ - 4x² - 28x – 16
x +2 is a factor
x=2
3
2
2
1
-8
5
-6
2
-5
-28 -16
-6
20
16
*Use only coefficients
3
-3
-4
-10 -8
0
no remainder
15 -21
f(x) (2 + x) (3x² - 10x -8)
7 -16
Remainder
f(x) (x+2) (3x+2) (x – 4)
finding the 0's of f(x)
(x+2) = -2
(3x+2) = -2/3
(x – 4) = 4
2x³ +x² - 8x +5 -16
x+3
Rachael SKinner
Factor
Solve
1.) Divide 2x⁴ + 3x³ + 5x – 1 by x² – 2x +2
Using Long Division
2.) Divide x³ + 2x² - 6x – 9 by x-2
Using Synthetic Division
3.) Factor Completely given that x-4 is a factor
f(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
-2
-6 20 -6
Synthetic Division
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)
f(x)=(x+2)(3x-1)(x-3)
x=-2,3,1/3
Factor the trinomial
and use the factor
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 a
n=4, a=0
• Evaluate the expression
Loren
Section 7.3
Power Functions and Function
Operations
Andy
Examples
1
2
1
2
Let f ( x)  4 x and g ( x)  9 x . Find the following.
a.) f ( x)  g ( x)
= 4x1/2 + (–9x1/2)
b.) f ( x)  g ( x)
= 4x1/2 – (–9x1/2)
1. Combine the coefficients1. Combine the
= [4 + (–9)]x1/2
= [4 – (–9)]x1/2
coefficients
2. Simplify
2. Simplify
= –5x1/2
= 13x1/2
Andy
Practice Problems
2
3
2
3
Let f ( x)  2 x and g ( x)  4 x . Find the following.
1. f (x) + g(x)
2.
f (x) – g(x)
Andy
Examples
3
4
Let f ( x)  6 x and g ( x)  x . Find the following.
a.
f (x)
f (x)
g(x)
b.
Plug in equation
= (6x)(x3/4)
Combine powers by
adding
= 6x(1 + 3/4)
Simplify
= 6x7/4
g(x)
Plug in equation
6x
= 3/4
x
Combine powers
by subtraction
= 6x(1 – 3/4)
Simplify
= 6x1/4
Andy
Practice Problems
1
5
Let f ( x)  4 x and g ( x)  x . Find the following.
4. f (x)
g(x)
5. f (x)
g(x)
Andy
Examples
Let f ( x)  4 x
a.
1
and g ( x)  5x  2. Find the following.
f(g(x))
Plug in equation
= f(5x – 2)
Plug g(x) into the equation
= 4(5x – 2)–1
g(f(x))
b.
Plug in equation
= g(4x–1)
Plug g(x) into the
equation
= 5(4x–1) – 2
Distribute the 5
Plug in the coefficient and power
of f(x)
= –20x–1 – 2
Simplify
20
= x –2
Andy
Examples
Let f ( x)  4 x
c.
1
and g ( x)  5x  2. Find the following.
f(f(x))
Plug in equation
= f(4x–1)
Plug f(x) into the equation
= 4(4x–1)–1
Distribute the exponent
= 4(4–1 x)
Distribute the 4
= 40x
OR
=x
Andy
Practice Problems
Let f ( x)  2 x  6 and g ( x)  3x 2 . Find the following.
a.
f(g(x))
b.
g(f(x))
c.
f(f(x))
Andy
7.4
Inverse Functions
Concepts
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).
Ileishka Ortiz
7.4
Original relation
Inverse relation
DOMAIN
DOMAIN
RANGE
RANGE
Ileishka Ortiz
1.) Find an equation for the inverse of the relation y
=2x–4
SOLUTION
y=2x–4
Write original relation.
x
Switch x and y
=2y – 4
x + 4 = 2y
Add 4 to each side.
1
x+2=y
2
Divide each side by 2.
Ileishka Ortiz
7.4
Functions f and g are inverses of each other provided:
f (g (x)) = x and
The function g is denoted by f
g ( f (x)) = x
–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.
Ileishka Ortiz
7.4
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.
f (g (x)) = f ( 1/2 x) + 2
= 2 (1/2 x ) + 2 – 4
= x+4 – 4
= x
Ileishka Ortiz
g (f (x)) = g (2x – 4)
= 1 (2x – 4) + 2
2
= x–2+2
= x
7.4
PRACTICE
Ileishka Ortiz
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 + k
y=a√x 0r y=a^3√x
H+K
H 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 WE
1. √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 + 1
DRAW THE GRAPH OF Y=-3√X
DASHED LINE IN GRAPH
IT 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 functions
From 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 - 1
1. 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 one
You 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
7.6
Powers 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 power
An extraneous solution is a trial solution that does not satisfy the original equation
Example
√x + 5 = 9
Isolate the radical by subtracting 5 by each side
(√x)^2 = 4^2
Simplify
x = 16
Radical Exponents
Example
3x^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 reciprocal
x = (64^1/3)4
Apply Properties of roots
x = 4^ 4 = 256
Simplify (The solution is 256) Check answer by substituting.
With Only One Radical
3√8x+3 - 5 = -2
Isolate the radical by adding 5 to each side
3√8x+3 = 3^3
Cube each side
8x +3 = 27
Subtract each side of by 3
8x=24
Simplify
x=3
Two Radicals
3√2x+4 = 2^3√3-x
Cube each side
(3√2x+4)^3 = (2^3√3-x)^3
Simplify
2x + 4 = 8(3-x)
Distribute
2x + 4 = 24 - 8x ( Add 8 x to each side
10x + 4 = 24
x=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 xintercepts
The 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 2
M=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 -1
Vertical asymptote is -4
M>n, so no horizontal
asymptote
Sydney
9.4
Multiplying and Dividing Rational
Expressions
Brett Robinson
• Let a, b, and c be nonzero real numbers or
variable expressions. Then the following
property applies:
• Divide out common Factor c
Brett Robinson
Multiplying
Factor
Numerator and
Denominator
Divide out common
factors before
multiplying if
possible
Simplified Form
Brett Robinson
Dividing
Multiply by the reciprocal
Factor
Divide out common
Factors
Simplified Form
Brett Robinson
Section 9.5
Addition, Subtraction, and Complex Fractions
Adding and Subtracting with Like
Denominators:
Solve the following problems:
Step I: Add numerators and simplify the expressions
4
5
4 5
9
3




3x
3x
3x
3x
x
QuickTime™ and a
decompressor
are needed to see this picture.
Step I: Subtract numerators
2x
4
2x  4


x  3
x  3
x  3
Jorge
Verde
Adding with Unlike Denominators
QuickTime™ and a
decompressor
are needed to see this picture.
Solution:
5
Step I: First find the LCD of 6 x 2 and 4 x
The LCD is 12 x
2
2
x
 12 x
( x  3)

 each expression.
Step 2: Use this to rewrite
5
x
5
x
5[2(x  3)]
x(3x)
 2




2
2
2
6x
4x  12x 6x
4x(x  3) 6x [2(x  3)] 4x(x  3)(3x)
10x  30
3x 2


2
12x (x  3) 12x 2 (x  3)
3x 2 10x  30

12x 2 (x  3)
Jorge
Verde
Subtracting with Unlike Denominators
QuickTime™ and a
decompressor
are needed to see this picture.
Solution
x 1
2
x 1
2
 2


2
2
x  4x  4
x 4
( x  2)
( x  2)( x  2)
( x  1)( x  2)
2(x  2)


2
(x  2) ( x  2) (x  2)( x  2)( x  2)
x 2  x  2  (2 x  4)

( x  2) 2 (x  2)
x 2  3x  6

(x  2) 2 ( x  2)
Jorge
Verde
Simplifying a Complex Fraction
QuickTime™ and a
decompressor
are needed to see this picture.
Solution
Step I: Add fractions in denominator.
Step 2: Multiply by reciprocal.
QuickTime™ and a
decompressor
are needed to see this pictur e.
Step 3: Divide out common factor.
Step 4: Write in simplified form.
Quick Time™a nd a
dec ompr esso r
ar e nee ded to see this pictur e.
QuickTime™ and a
decompressor
are needed to see this picture.
Qu ic kT i me™ a nd a
de co mpre ss or
are n ee de d t o s ee t h is pi ct u r
e.
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.
Morgan Hillhouse
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.
Cross Multiplying
X
―
X-4
=
3
―
4
4X=3X-12
-3X
4X=-12
4
X=-3
Morgan Hillhouse
Continued Problem #2
X+6
――
X
=
Morgan Hillhouse
7
―
8
8X+48=7X
-48 -7X
X=-48
Multiple Rational Expressions
2X
――
X-2
-
(3X+2)
(X-2)
Morgan Hillhouse
4X
17X+4
―― = ――――
3X+2
3X₂-4X-4
2X
――
X-2
(3X+2)
(X-2)
-
4X
17X+4
―― = ――――
3X+2
(3X+2)(X-2)
CONTINUED ON
NEXT SLIDE
Continued
2X(3X+2)-(4X-1)(X-2)=17X+4
6X₂+4X-4X₂+9X-2=17+4
-17-4
2X₂-4X-6=0
6X₂+4X-(4X₂-9X+2)=17X+4
Morgan Hillhouse
2(X₂-2X-3)=0
2(X-3)(X+1)=0
X-3=0 X+1=0
X=3 or -1