Review Topics (Chapter 0 & 1)
• Exponents & Radical Expressions
• Factoring
• Quadratic Equations
• Rational Expressions
• Rational Equations & Clearing Fractions
• Radical Equations
• Solving Inequalities
• Linear Graphing and Functions
• Function Evaluation
• Slope and Average Rate of Change
• Difference Quotient
Review of Exponents
82 =8 • 8 = 64
24 = 2 • 2 • 2 • 2 = 16
x2 = x • x
Base = x
Exponent = 2
x4 = x • x • x • x
Base = x
Exponent = 4
Exponents of 1
Anything to the 1 power is itself
51 = 5
Zero Exponents
Anything to the zero power = 1
x1 = x (xy)1 = xy
50 = 1
x0 = 1
(xy)0 = 1
Negative Exponents
5-2 = 1/(52) = 1/25
x-2 = 1/(x2)
xy-3 = x/(y3)
a-n = 1/an
1/a-n = an
a-n/a-m = am/an
(xy)-3 = 1/(xy)3 = 1/(x3y3)
Raising Quotients to Powers
a
b
n
Examples:
an
= bn
-n
a
b
2
3
4
32
42
=
2x
y
2x
y
3
=
=
bn
an
= b
a
9
= 16
3
(2x)
= 3 =
y
-3
=
a-n
b-n
(2x)-3
=
y-3
8x3
y3
1
y-3(2x)3
=
y3
y3
(2x)3 = 8x3
n
Product Rule
am • an = a(m+n)
x3 • x5 = xxx • xxxxx = x8
x-3 • x5 = xxxxx = x2 = x2
xxx
1
x4 y3 x-3 y6 = xxxx•yyy•yyyyyy = xy9
xxx
3x2 y4 x-5 • 7x = 3xxyyyy • 7x = 21x-2 y4 = 21y4
xxxxx
x2
Quotient Rule
am = a(m-n)
an
43 = 4 • 4 • 4 = 41 = 4
42
4•4
43 = 64
42
16
x5 = xxxxx = x3
x2
xx
x5 = x(5-2) = x3
x2
15x2y3 = 15 xx yyy = 3y2
5x4y
5 xxxx y
x2
3a-2 b5 = 3 bbbbb bbb = b8
9a4b-3
9aaaa aa
3a6
= 8
2
= 4
15x2y3 = 3 • x -2 • y2 = 3y2
5x4y
x2
3a-2 b5
9a4b-3
= a(-2-4)b(5-(-3)) = a-6 b8 = b8
3
3
3a6
Powers to Powers
(am)n = amn
(a2)3 a2 • a2 • a2 = aa aa aa = a6
(24)-2 = 1
(24)2
(x3)-2 = x –6
(x -5)2 x –10
= 1
=1
= 1/256
24 • 24 16 • 16
= x 10 = x4
x6
(24)-2 = 2-8 = 1 = 1
28 256
Products to Powers
(ab)n = anbn
(6y)2 = 62y2 = 36y2
(2a2b-3)2 = 22a4b-6 = 4a4
= a
4(ab3)3
4a3b9
4a3b9b6
b15
What about this problem?
5.2 x 1014
3.8 x 105
= 5.2/3.8 x 109  1.37 x 109
Do you know how to do exponents on the calculator?
Square Roots & Cube Roots
A number b is a square root
of a number a if b2 = a
A number b is a cube root
of a number a if b3 = a
25 = 5 since 52 = 25
3
Notice that 25 breaks down into 5 • 5
So, 25 =  5 • 5
Notice that 8 breaks down into
3
2 • 2 • 2 So, 8 =  2 • 2 • 2
See a ‘group of 2’ -> bring it outside the
radical (square root sign).
See a ‘group of 3’ –> bring it outside
the radical (the cube root sign)
Example: 200 = 2 • 100
= 2 • 10 • 10
= 10 2
Note: -25 is not a real number since no
number multiplied by itself will be negative
8 = 2 since 23 = 8
3
3
Example: 200 = 2 • 100
3
= 2 • 10 • 10
3
= 2 • 5 • 2 • 5 • 2
3
= 2 • 2 • 2 • 5 • 5
3
= 2 25
3
Note: -8
IS a real number (-2) since
-2 • -2 • -2 = -8
Nth Root ‘Sign’ Examples
Even radicals of positive numbers
Have 2 roots. The principal root
Is positive.

16 = 4 or -4

-16 not a real number

-16 not a real number

-32
= -2
Odd radicals of negative numbers
Have 1 negative root.

32
= 2
Odd radicals of positive numbers
Have 1 positive root.
4
5
5
Even radicals of negative numbers
Are not real numbers.
Exponent Rules
(x )  x
m n
mn
x 1
0
(XY)m = xmym
x x  x
m
n
m n
m
x
m n

x
n
x
m
X
Y
Xm
= m
Y
x
m
m
1
 m
x
x x
1/ m
Examples to Work through
3
4
27 
81 
8
8
5/ 4
1/ 4
12 
3
8x4 y3 
8
3/ 4

Some Rules for Simplifying
Radical Expressions
n
a  b  ab
n
a a
n
a a
n
m
n
1/ n
m/n
Practice Problems
3
5
5 7 
3
3
16 
5t  3 125t 
6
2 x 5 16 y


y
x
300 
3
54 
4
512 x 
4
5
4t 5 8t
 6 
r
r
5
4 5 8


 9  27
Operations on Radical Expressions
•Addition and Subtraction (Combining LIKE Terms)
•Multiplication and Division
• Rationalizing the Denominator
Radical Operations with
Numbers
3 2 4 2 
3
8 x 23 x


27 3
4
xy  x y 
23 16  53 54  103 2 
5
4
5
2 3z  3 12 z  3 48 z 
Multiplying Radicals (FOIL
works with Radicals Too!)
( 2 x  3 y )( 2 x  3 y ) 
( x  9)( x  8) 
Rationalizing the Denominator
• Remove all radicals from the denominator
1

2
1

32
xy
3

32
y
3

Adding & Subtracting Polynomials
Combine Like Terms
(2x2 –3x +7) + (3x2 + 4x – 2) =
5x2 + x + 5
(5x2 –6x + 1) – (-5x2 + 3x – 5) = (5x2 –6x + 1) + (5x2 - 3x + 5)
= 10x2 – 9x + 6
Types of Polynomials
f(x) = 3
f(x) = 5x –3
f(x) = x2 –2x –1
f(x) = 3x3 + 2x2 – 6
Degree 0
Degree 1
Degree 2
Degree 3
Constant Function
Linear
Quadratic
Cubic
Multiplication of Polynomials
Step 1: Using the distributive property, multiply every term in the
1st polynomial by every term in the 2nd polynomial
Step 2: Combine Like Terms
Step 3: Place in Decreasing Order of Exponent
4x2 (2x3 + 10x2 – 2x – 5) = 8x5 + 40x4 –8x3 –20x2
(x + 5) (2x3 + 10x2 – 2x – 5) = 2x4 + 10x3 – 2x2 – 5x
+ 10x3 + 50x2 – 10x – 25
= 2x4 + 20x3 + 48x2 –15x -25
Binomial Multiplication with FOIL
(2x + 3) (x - 7)
F.
(First)
O.
(Outside)
I.
(Inside)
L.
(Last)
(2x)(x)
(2x)(-7)
(3)(x)
(3)(-7)
3x
-21
2x2
-14x
2x2
-14x
2x2
+
- 11x
3x
-21
-21
Division by a Monomial
3x2 + x
x
4x2 + 8x – 12
4x2
15A2 – 8A2 + 12
4A
5x3 – 15x2
15x
5x2y + 10xy2
5xy
12A5 – 8A2 + 12
4A
Review: Factoring Polynomials
To factor a number such as 10, find
out
‘what times what’ = 10
10 = 5(2)
To factor a polynomial, follow a similar process.
Factor: 3x4 – 9x3 +12x2
3x2 (x2 – 3x + 4)
Another Example:
Factor 2x(x + 1) + 3 (x + 1)
(x + 1)(2x + 3)
Solving Polynomial Equations By
Factoring
Zero Product Property : If AB = 0 then A = 0 or B = 0
Solve the Equation: 2x2 + x = 0
Step 1: Factor
x (2x + 1) = 0
Step 2: Zero Product
x = 0 or
2x + 1 = 0
Step 3: Solve for X
x = 0 or
x= -½
Question: Why are there 2 values for x???
Factoring Trinomials
To factor a trinomial means to find 2 binomials whose product
gives you the trinomial back again.
Consider the expression: x2 – 7x + 10
The factored form is:
(x – 5) (x – 2)
Using FOIL, you can multiply the 2 binomials and
see that the product gives you the original trinomial expression.
How to find the factors of a trinomial:
Step 1: Write down 2 parentheses pairs.
Step 2: Do the FIRSTS
Step3 : Do the SIGNS
Step4: Generate factor pairs for LASTS
Step5: Use trial and error and check with FOIL
Practice
Factor:
1. y2 + 7y –30
2.
10x2 +3x –18
3. 8k2 + 34k +35
4. –15a2 –70a + 120
5. 3m4 + 6m3 –27m2
6. x2 + 10x + 25
Special Types of Factoring
Square Minus a Square
A2 – B2 = (A + B) (A – B)
Cube minus Cube and Cube plus a Cube
(A3 – B3) = (A – B) (A2 + AB + B2)
(A3 + B3) = (A + B) (A2 - AB + B2)
Perfect Squares
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Quadratic Equations
General Form of Quadratic Equation
ax2 + bx + c = 0
a, b, c are real numbers & a 0
A quadratic Equation: x2 – 7x + 10 = 0
a = _____
1 b = _____
-7 c = ______
10
Methods & Tools for Solving Quadratic Equations
1. Factor
2. Apply zero product principle (If AB = 0 then A = 0 or B = 0)
3. Square root method
Example1:
Example 2:
4. Completing the Square x2 – 7x + 10 = 0
4x2 – 2x = 0
5. Quadratic Formula
(x – 5) (x – 2) = 0
2x (2x –1) = 0
x – 5 = 0 or x – 2 = 0
2x=0 or 2x-1=0
+5 +5
+2 +2
2 2
+1 +1
2x=1
x = 5
or
x= 2
x = 0 or x=1/2
Square Root Method
If u2 = d then u = d or u = - d. If u2 = d then u = + d
Solving a Quadratic Equation with the Square Root Method
Example 1:
Example 2:
4x2 = 20
(x – 2)2 = 6
4
4
x – 2 = +6
x2 = 5
+2 +2
x = + 5
So, x = 5 or - 5
x = 2 + 6
So, x = 2 + 6 or 2 - 6
Completing the Square
(Example 1)
If x2 + bx is a binomial then by adding b 2 which is the square of half
2
the coefficient of x, a perfect square trinomial results:
x2 + bx + b 2 = x + b 2
2
2
Solving a quadratic equation with ‘completing the square’ method.
Example:
x2 - 6x + 2 = 0
-2 -2
x2 - 6x = -2
x2 - 6x + 9 = -2 + 9
(x – 3)2 = 7
x – 3 = + 7
x = (3 + 7 ) or (3 - 7 )
Step1: Isolate the Binomial
Step 2: Find ½ the coefficient of x (-3 )
and square it (9) & add to both sides.
Note: If the coefficient of x2 is not 1 you
must divide by the coefficient of x2 before
completing the square. ex: 3x2 – 2x –4 = 0
(Must divide by 3 before taking ½ coefficient of x)
Step 3: Apply square root method
(Completing the Square – Example 2)
Step 1: Check the coefficient of the x2 term. If 1 goto step 2
If not 1, divide both sides by the coefficient of the x2 term.
Step 2: Calculate the value of : (b/2)2
[In this example: (2/2)2 = (1)2 = 1]
Step 3: Isolate the binomial by grouping the x2 and x term together, then
add (b/2)2 to both sides of he equation.
2 +4x – 1 = 0
2x
Step 4: Factor & apply square root method
2x2 +4x – 1 = 0
2
2 2 2
(x + 1) (x + 1) = 3/2
(x + 1)2 = 3/2
x2 +2x – 1/2 = 0
(x2 +2x
)=½
(x2 +2x + 1 ) = 1/2 + 1
√(x + 1)2 = √3/2
x + 1 = +/- √6/2
x = √6/2 – 1 or - √6/2 - 1
Quadratic Formula
General Form of Quadratic Equation:
ax2 + bx + c = 0
Quadratic Formula: x = -b + b2 – 4ac
2a
discriminant: b2 – 4ac
if 0, one real solution
if >0, two unequal real solutions
if <0, imaginary solutions
Solving a quadratic equation with the ‘Quadratic Formula’
2x2 – 6x + 1= 0
_______
2
a = ______
-6
b = ______
x = - (-6) + (-6)2 – 4(2)(1)
2(2)
= 6 + 36 –8
4
= 6 + 28
4
= 6 + 27
4
= 2 (3 + 7 ) = (3 + 7 )
4
2
c=
1
Solving Higher Degree Equations
x3 = 4x
2x3 + 2x2 - 12x = 0
x3 - 4x = 0
x (x2 – 4) = 0
x (x – 2)(x + 2) = 0
2x (x2 + x – 6) = 0
x=0
2x (x + 3) (x – 2) = 0
x–2=0 x+2=0
2x = 0 or
x + 3 = 0 or x – 2 = 0
x=2
x=0
x = -3 or
x = -2
or
x=2
Solving By Grouping
x3 – 5x2 – x + 5 = 0
(x3 – 5x2) + (-x + 5) = 0
x2 (x – 5) – 1 (x – 5) = 0
(x – 5)(x2 – 1) = 0
(x – 5)(x – 1) (x + 1) = 0
x–5=0
or
x-1=0
or
x+1=0
x=5
or
x=1
or
x = -1
Rational Expressions
Rational Expression – an expression in which a polynomial is divided by
another nonzero polynomial.
Examples of rational expressions
4
x
x
2x – 5
Domain = {x | x  0}
Domain = {x | x  5/2}
2
x–5
Domain = {x | x  5}
Multiplication and Division of
Rational Expressions
A • C = A
B • C
B
9x
3x2
5y – 10 = 5 (y – 2)
10y - 20
10 (y – 2)
2z2 – 3z – 9
z2 + 2z – 15
A2 – B2
A+B
3
x
= 5 =1
10
2
= (2z + 3) (z – 3)
(z + 5) (z – 3)
=
=
(A + B)(A – B)
(A + B)
= 2z + 3
z+5
= (A – B)
Negation/Multiplying by –1
-y – 2
4y + 8
=
y+2
4y + 8
OR
-y - 2
-4y - 8
Examples
x3 – x
x–1
(x3 – x) (x + 1)
=
x(x – 1)
= x (x2 – 1)(x + 1)
x(x – 1)
= x (x + 1) (x – 1)(x + 1)
x(x – 1)
=
x2 – 25
x2 –10x + 25

2
x + 5x + 4
2x2 + 8x
• x+1
x
(x + 1)(x + 1) = (x + 1)2
=
x2 – 25
x2 + 5x + 4
•
2x2 + 8x
x2 –10x + 25
(x + 5) (x – 5) • 2x(x + 4)
= (x + 4)(x + 1) • (x – 5) (x – 5)
2x (x + 5)
= (x + 1)(x – 5)
Check Your Understanding
Simplify:
Simplify:
x2 –6x –7
x2 -1
1
x-2
6
(x + 1) (x –7)
(x + 1) (x – 1)
(x – 7)
(x – 1)
1
x–2
1
x–2

3
x2 + x x2 + x - 6
3
•
•
(x + 3)
3
(x + 3) (x – 2)
3
Addition of Rational Expressions
Adding rational expressions is like adding fractions
With LIKE denominators:
1
8
+
x
x+2
x
3x2 + 4x - 4
2
8
+
+
=
3x - 1
x+2
=
3
8
4x - 1
x+2
2
(2 + x)
(2 + x)
=
=
3x2 + 4x -4
(3x2 + 4x – 4) (3x -2)(x + 2)
=
1
(3x – 2)
Adding with UN-Like Denominators
3
4
1
x2 – 9
+ 1
8
(3) (2)
8
+ 1
8
6
8
1
8
+
7
8
+
2
x+3
1
(x + 3)(x – 3)
+
2
(x + 3)
1
(x + 3)(x – 3)
+
2 (x – 3)
(x + 3)(x – 3)
1 + 2(x – 3)
(x + 3) (x – 3)
= 1 + 2x – 6
(x + 3) (x – 3)
= 2x - 5
(x + 3) (x – 3)
Subtraction of Rational Expressions
To subtract rational expressions:
Step 1: Get a Common Denominator
Step 2: Combine Fractions DISTRIBUTING the ‘negative sign’
BE CAREFUL!!
2x
x2 – 1
=
-
x–1
(x + 1)(x –1)
x+1
x2 - 1
=
=
2x – (x + 1)
x2 -1
1
(x + 1)
= 2x2 – x - 1
x -1
Check Your Understanding
Simplify:
b
2b - 4
b-1
b-2
-
b
2(b – 2)
-
b-1
b-2
b
2(b – 2)
+
-b+1
b-2
b
2(b – 2)
+ 2(-b+1)
2(b – 2)
b –2b+2
2(b – 2)
=
-b + 2
2(b – 2)
=
-1(b – 2)
2(b – 2)
=
-1
2
Complex Fractions
A complex fraction is a rational expression that contains
fractions in its numerator, denominator, or both.
Examples:
1
5
x
x2 – 16
1
x
+
2
x2
4
7
1
x-4
3
x
-
1
x2
7/20
x
x+4
x+2
3x - 1
Rational Equations
(2x – 1)
3x
= 3
2x – 1
3x = 3(2x – 1)
3x = 6x – 3
-3x = -3
(x - 2)
x+1
x–2
(x + 1)
=
3
x-2
6
x+1
=x
x+1=3
6 = x (x + 1)
x=2
6 = x2 + x
x=1
x2 + x – 6 = 0
(x + 3 ) (x - 2 ) = 0
Careful! – What do
You notice about
the answer?
x = -3
or
x=2
Rational Equations Cont…
To solve a rational equation:
Step 1: Factor all polynomials
Step 2: Find the common denominator
Step 3: Multiply all terms by the common denominator
Step 4: Solve
(12x)
x+1 - x–1 = 1
2x
4x
3
=
6 (x + 1) -3(x – 1) = 4x
6x + 6 –3x + 3 = 4x
3x + 9 = 4x
-3x
-3x
9 = x
Other Rational Equation Examples
3
x–2
(x + 2)(x – 2)
3
x–2
(4x2)
+
5
x+2
=
12
x2 - 4
1 + 1 = 3
x
x2
4
+
5
x+2
=
12
(x + 2) (x – 2)
4x + 4 = 3x2
3x2 - 4x - 4 = 0
3(x + 2)
+
5(x – 2) =
3x + 6
+
5x – 10 =
8x – 4 = 12
+4 +4
8x
= 16
x
= 2
12
12
(3x + 2) (x – 2) = 0
3x + 2 = 0 or x – 2 = 0
3x = -2
or
x=2
x = -2/3
or
x=2
Check Your Understanding
Simplify:
x
1
+
x2 – 1
x2 – 1
1
x–2
1
x(x – 1)
-
1
x-1
2(x – 3)
x(x – 2)
3
x
+ 1
x2 – 1
-
2
x(x + 1)
Solve
6 - 1
x
2
4
=1
3
= 2
2x – 1
x+1
2
+ 3
x–1
x+2
5
=
x
x2 + x - 2
-1/4
3
x(x – 1)(x + 1)
Try this one:
Solve for p:
1 =1 + 1
F
p
q
Radical Equations Continued…
Example1:
Example 2:
x + 26 – 11x = 4
26 – 11x = 4 - x
X2 = 64
(26 – 11x)2 = (4 – x)2
26 – 11x = (4-x) (4-x)
26 - 11x = 16 –4x –4x +x2
26 –11x = 16 –8x +
-26 +11x -26 +11x
0 = x2 + 3x -10
0 = (x - 2) (x + 5)
x – 2 = 0 or x + 5 = 0
x=2
x = -5
x2
Example 3:
(3x  6)  25
2
Inequality Set & Interval Notation
Set Builder Notation
{1,5,6}
{ }

{x | x > -4}
x such that
x is greater than –4
{x | x < 2}
x such that x is less
than or equal to 2
Interval (-4, )
Notation
Graph
-4
{6}
{x | -2 < x < 7}
x such that x is greater
than –2 and less than
or equal to 7
(-, 2]
(-2, 7]
2
-2
0
Question: How would you write the set of all real numbers?
7
(-, ) or
R
Inequality Example
Statement
7x + 15 > 13x + 51
Reason
[Given]
-6x + 15 > 51
[-13x]
-6x > 36
[-15]
x
< 6
[Divide by –6, so must ‘flip’ the inequality sign
Set Notation: {x | x < 6}
Interval Notation: (-, 6]
Graph:
6
Graphs
y axis
Quadrant II (-, +) Quadrant I (+, +)
y - $$ in thousands
Origin (0, 0)
-6
-2
(-6,-3)
(6,0)
x axis
2 4 6
(5,-2)
y
intercept
(0,-3)
x
Yrs
x
intercept
Quadrant III (-, -) Quadrant IV (+, -)
When distinct points are plotted as above Graphs represent trends in data.
the graph is called a scatter plot – ‘points For example:
x – number of years in business
that are scattered about’
y – thousands of dollars of profit
A point in the x/y coordinate plane is
Equation : y = ½ x – 3
described by an ordered pair of
coordinates (x, y)
Linear Equations
The graph of a linear equation is a line.
A linear function is of the form y = mx + b, where m and b are constants.
y
y = 3x + 2
y = 3x + 5x
y = -2x –3
x
y = (2/3)x -1
y=4
6x + 3y = 12
All of these equations are linear.
Three of them are graphed above.
x y=3x+2
x y=2/3x –1
0 2
0
-1
1 5
3
1
X and Y intercepts
Equation: y = ½ x – 3
y
(6,0)
y
intercept
(0,-3)
x
x
intercept
The y intercept happens where y is something & x = 0: (0,-3____)
Let x = 0 and solve for y: y = ½ (0) – 3 = -3
The x intercept happens where x is something & y = 0:6(____, 0)
Let y = 0 and solve for x: 0 = ½ x – 3 => 3 = ½ x
=> x = 6
Slope
Slope is the ratio of
y 2 – y1
RISE (How High)
=
RUN (How Far)
y
x2 – x1
y (Change in y)
x (Change in x)
Slope = 5 – 2 = 3
1-0
Slope = 1 – (-1) = 2
3–0
3
Things to know:
1. Find slope from graph
2. Find a point using slope
3. Find slope using 2 points
4. Understand slope between
2 points is always the same
on the same line
x y = mx + b
m = slope
b = y intercept
y=2/3x –1
x y=3x+2
x
0 2
0
-1
1 5
3
1
The Possibilities for a Line’s Slope (m)
Negative Slope
Positive Slope
y
Zero Slope
y
m>0
Undefined Slope
y
m<0
y
m is
undefined
m=0
x
x
Line rises from left to
right.
Line falls from left to
right.
Line is horizontal.
Example:
y = -½ x + 1
Example:
y=2
Example:
y=½x+2
x
x
Line is vertical.
Example:
x=3
Question: If 2 lines are parallel do you know anything about their slopes?
Things to know:
1. Identify the type of slope given a graph.
2. Given a slope, understand what the graph would look like and draw it.
3. Find the equation of a horizontal or vertical line given a graph.
4. Graph a horizontal or vertical line given an equation
5. Estimate the point of the y-intercept or x-intercept from a graph.
Linear Equation Forms (2 Vars)
Standard Form
Ax + By = C
Example: 6x + 3y = 12
Slope Intercept Form
Example: y = - ½ x - 2
Point Slope Form
A, B, C are real numbers.
A & B are not both 0.
Things to know:
1. Graph using x/y chart
2. Know this makes a line graph.
y = mx + b
m is the slope
b is the y intercept
Things to know:
1. Find Slope & y-intercept
2. Graph using slope & y-intercept
3. Application meaning of of slope & intercepts
y – y1 = m(x – x1)
Example: Write the linear equation through point P(-1, 4) with slope 3
y – y1 = m(x – x1)
Things to know:
y – 4 = 3(x - - 1)
1. Change from point slope to/from other forms.
y – 4 = 3(x + 1)
2. Find the x or y-intercept of any linear equation
Practice Problems
1.
Find the slope of a line passing through (-1, 2) and (3, 8)
2.
Graph the line passing through (1, 2) with slope of - ½
3.
Is the point (2, -1) on the line specified by: y = -2(x-1) + 3 ?
1.
Find the equation of a line with slope = 4 through the point (-1,5)
1.
Find the equation of a line passing through the points (-2, 1) and (3, 7)
8. Graph (using an x/y chart – plotting points) and find intercepts of
any equation such as: y = 2x + 5 or y = x2 – 4
A Rational Function Graph
y= 1
x
x
-2
-1
-1/2
0
½
1
2
y
-1/2
-1
-2
Undefined
2
1
½
Intercepts:
No intercepts exist
If y = 0, there is no solution for x.
If x = 0, y is undefined
The line x = 0 is called a vertical asymptote.
The line y = 0 is called a horizontal asymptote.
Functions and Graphs
Year
1997
1998 1999 2000
$3111 $3247 $3356 $3510
Cost
The cost depends on the year.
independent variable (x)
dependent variable (y)
The table above establishes a relation between the year and the cost of tuition
at a public college. For each year there is a cost, forming a set of ordered pairs.
A relation is a set of ordered pairs (x, y). The relation above can be written
as 4 ordered pairs as follows:
S = {(1997, 3111), (1998, 3247), (1999, 3356), (2000, 3510)}
x
y
x
y
x
y
x
y
Domain – the set of all x-values. D = {1977, 1998, 1999, 2000}
Range – the set of all y-values. R = {3111, 3247, 3356, 3510}
Thinking Exercise: Draw a ‘line’ in the x/y axes.
What is the Domain & Range?
Year(x)
1997
1998
1999
2000
Cost(y)
3111
3247
3356
3510
Input
x
Functions & Linear Data Modeling
y – Profit in thousands of $$
(Dependent Var)
Function
f
(6,0)
Output
y=f(x)
y
intercept
(0,-3)
x - Years in business
(Independent Var)
x
intercept
Equation: y = ½ x – 3
Function: f(x) = ½ x – 3
A function has exactly one output value (y)
for each valid input (x).
x
0
2
6
8
Use the vertical line test to see if an equation is
a function.
•If it touches 1 point at a time then FUNCTION
•If it touches more than 1 point at a time then
NOT A FUNCTION.
y = f(x)
-3 f(0) = ½(0)-3=-3
-2 f(2) = ½(2)-3=-2
0 f(6) = ½(6)-3=0
1 f(8) = ½(8)-3=1
How to Determine if an equation is a
function
Graphically: Use the vertical line test
Symbolically/Algebraically: Solve for y to see if there is only 1 y-value.
Example 1: x2 + y = 4
Example 2: x2 + y2 = 4
y = 4 – x2
y2 = 4 – x2
For every value of x there
y = 4 – x2 or y = - 4 – x2
Is exactly 1 value for y, so
This equation IS A FUNCTION. For every value of x there
are 2 possible values for y, so
This equation IS NOT A
FUNCTION.
Are these graphs functions?
Use the vertical line test to tell if the following are functions:
y = x2
y = x3
Origin
Symmetry
Y-axis
Symmetry
x = y2
X-axis
Symmetry
More on Evaluation of Functions
f(x) = x2 + 3x + 5
Evaluate: f(2)
f(2) = (2)2 + 3(2) + 5
f(2) = 4 + 6 + 5
f(2) = 15
Evaluate: f(x + 3)
f(x + 3) = (x + 3)2 + 3 (x + 3) + 5
f(x + 3) = (x + 3)(x + 3) + 3x + 9 + 5
f(x + 3) = (x2 + 3x + 3x + 9) + 3x + 14
f(x + 3) = (x2 + 6x + 9) + 3x + 14
f(x + 3) = x2 + 9x + 23
Evaluate: f(-x)
f (-x) = ( -x)2 + 3( -x) + 5
f (-x) = x2 - 3x + 5
More on Domain of Functions
A function’s domain is the largest set of real numbers for which the value f(x)
is a real number. So, a function’s domain is the set of all real numbers
MINUS the following conditions:
• specific conditions/restrictions placed on the function
• Bounds relating to real-life data modeling
(Example: y = 7x, where y is dog years and x is dog’s age)
• values that cause division by zero
• values that result in an even root of a negative number
What is the domain the following functions:
1.
f(x) = 6x
x2 – 9
2. g(x) =
3x + 12
3. h(x) = 2x + 1
Slope & Average Rate of Change
y - $$ in thousands
y=½x–3
(6,0)
y = x2 - 4x + 4
x
Yrs
(0,-3)
The slope of a line may be
interpreted as the rate of change.
The rate of change for a line is
constant (the same for any 2 points)
y2 – y1
x2 – x1
Non-linear equations do not have a
constant rate of change. But you can
Find the average rate of change from
x1 to x2 along a secant to the graph.
f(x2) – f(x1)
x2 – x1
Definition of a Difference Quotient
The average rate of change for f(x) is called the “difference quotient”
and is defined below. (This is an important concept in calculus – it
becomes the mathematical definition of the derivative you will learn
About this semester. f (x + h) - f (x)
h
Example: Find the difference quotient for : f(x) = 2x2 -3
f(x + h) = 2(x + h )2 - 3
= 2(x + h)(x + h) -3
= 2(x2 + 2xh + h2 ) -3
= 2x2 + 4xh + 2h2 -3
=
=
=
So, the difference quotient is:
2x2 + 4xh + 2h2 -3 – (2x2 -3)
h
2x2 + 4xh + 2h2 -3 – 2x2 + 3
h
4xh + 2h2
h
4x + 2h
Increasing, Decreasing, and Constant
Functions
A function is increasing on an interval if for any x1, and x2 in the
interval, where x1 < x2, then f (x1) < f (x2).
A function is decreasing on an interval if for any x1, and x2 in the
interval, where x1 < x2, then f (x1) > f (x2).
A function is constant on an interval if for any x1, and x2 in the
interval, where x1 < x2, then f (x1) = f (x2).
(x2, f (x2))
(x1, f (x1))
(x2, f (x2))
(x1, f (x1))
(x1, f (x1))
(x2, f (x2))
Increasing
f (x1) < f (x2)
Decreasing
f (x1) > f (x2)
Constant
f (x1) = f (x2)
More Examples
a.
b.
5
5
4
4
3
3
2
1
1
-5 -4 -3 -2
-1
-1
-2
1
2
3 4
5
-5 -4 -3 -2
-1
-1
-2
-3
-3
-4
-5
-4
-5
Observations
• Decreasing on the
interval (-oo, 0)
1
2
3 4
5
Observations
a. Two pieces (a piecewise function)
b. Constant on the interval (-oo, 0).
•
•
Increasing on the
interval (0, 2)
Decreasing on the
interval (2, oo).
c. Increasing on the interval (0, oo).
Challenge Yourself: What might be
the definition of the piecewise function
for this graph? (You will learn about these
Later. Can you guess what it might be?)
f(x) = sin (x) x
0
/2
The point at which a function changes its increasing or decreasing

behavior is called a relative minimum or relative maximum.
3/2
y
2
2
Relative (local) Min & Max
(90, f(90))
y
0
1
0
-1
0
f(90), or 1, is a local max
1
x
0
90
180
270
360
-1
-2
A function value f(a) is a relative
maximum of f if there exists an open
interval about a such that f(a) > f(x)
for all x in the open interval.
(270, f(270))
f(270), or -1, is a local min
A function value f(b) is a relative
minimum of f if there exists an open
interval about b such that f(b) < f(x)
for all x in the open interval.
Library of Functions/Common Graphs
y=c
y=x
y = x2
x
x
y = x3
y=
x
y = |x|
x
y = 1/x
x
x
y = x1/3
x
x
Piecewise Functions
A function that is defined by two (or more) equations over
a specified domain is called a piecewise function.
f(x) =
x2 + 3
5x + 3
if x < 0
if x>=0
f(-5) = (-5)2 + 3 = 25 + 3 = 28
f(6) = 5(6) + 3 = 33
See Page 247 for more examples
Sum, Difference, Product, and Quotient
of Functions
Let f and g be two functions. The sum of f + g, the difference f – g,
the product fg, and the quotient f /g are functions whose domains are the set of
all real numbers common to the domains of f and g, defined as follows:
Sum:
Difference:
Product:
Quotient:
(f + g)(x) = f (x)+g(x)
(f – g)(x) = f (x) – g(x)
(f • g)(x) = f (x) • g(x)
(f / g)(x) = f (x)/g(x), provided g(x) does not equal 0
Example: Let f(x) = 2x+1 and g(x) = x2-2.
f+g = 2x+1 + x2-2 = x2+2x-1
f-g = (2x+1) - (x2-2)= -x2+2x+3
fg = (2x+1)(x2-2) = 2x3+x2-4x-2
f/g = (2x+1)/(x2-2)
Adding & Subtracting Functions
If f(x) and g(x) are functions, then:
(f + g)(x) = f(x) + g(x)
(f – g)(x) = f(x) – g(x)
Examples: f(x) = 2x + 1 and g(x) = -3x – 7
Method1
Method1
(f + g)(4) = 2(4) + 1 + -3(4) – 7
(f – g)(6) = 2(6) + 1 – [-3(6) – 7]
= 8+1
+ -12 – 7
= 12 + 1 - [-18 – 7]
= 9
+ -19
= 13 - [-25]
=
-10
= 13 + 25
Method2
Method2 =
38
(f + g)(4) = 2x + 1 + -3x – 7
(f - g)(6) = 2x + 1 - [-3x – 7]
=
-x – 6
= 2x + 1 + 3x + 7
=
-4 – 6
= 5x + 8
= - 10
= 5(6) + 8
= 30 + 8
=
38
Adding/subtracting also extends to non-linear
functions you will see in a subsequent chapter.
Exponential Functions
Exponential function
– any function whose equation contains a variable in the exponent.
[measures rapid increase or decrease (Example: epidemic growth)]
f(x) = bx
f – exponential function b - constant base (b > 0, b  1)
number -- domain is (-∞ , ∞)
f(x) = 2x
g(x) = 10x
Graphing Exponential Functions:
x
f (x) = 3x
g(x) = 3x+1
-2
3-2 = 1/9
3-2+1 = 3-1 = 1/3
-1
3-1 = 1/3
3-1+1 = 30 = 1
0
30 = 1
30+1 = 31 = 3
1
31 = 3
31+1 = 32 = 9
2
32
32+1
h(x) = 3x+1
g(x) = 3x+1
f (x) = 3x
x = any real
Shift up c units
f(x) = bx + c
Shift down c units
f(x) = bx – c
Shift left c units
f(x) = bx+c
(-1, 1)
(0, 1)
=9
=
33
= 27
-5 -4 -3 -2 -1
1 2 3 4 5 6
Shift right c units
f(x) = bx-c
The Natural Base e
An irrational number, symbolized by
the letter e, appears as the base in
many applied exponential functions.
This irrational number is
approximately equal to 2.72. More
accurately,
The number e is called the natural
base. The function f (x) = ex is called
the natural exponential function.
f (x) = 3x f (x) = ex
4
f (x) = 2x
(1, 3)
3
(1, e)
2
(1, 2)
(0, 1)
-1
1
Logarithmic Functions
A logarithm is an exponent such that for b > 0, b  1 and x > 0
y = logb x
if and only if
by = x
Logarithmic equations
1) 2 = log5 x
2) 3 = logb 64
3) log3 7 = y
4) y = loge 9
Corresponding exponential forms
1) 52 = x
2) b3 = 64
3) 3y = 7
4) ey = 9
Evaluate the Question Needed for
Logarithmic Evaluation
Expression
Logarithmic Expression
Evaluated
log2 16
2 to what power is 16?
log2 16 = 4 because 24 = 16.
log3 9
3 to what power is 9?
log3 9 = 2 because 32 = 9.
log25 5
25 to what power is 5?
log25 5 = 1/2 because 251/2 = 5.
Logarithmic Properties
Logb b = 1
Logb 1 = 0
1 is the exponent to which b must be raised to obtain b. (b1 = b).
0 is the exponent to which b must be raised to obtain 1. (b0 = 1).
logb bx = x
b logb x = x
The logarithm with base b of b raised to a power equals that powe
b raised to the logarithm with base b of a number equals that num
Graphs of f (x) = 2x and g(x) = log2 x [Logarithm is the inverse of the exponential
function]x
-2
-1
0
1
2
3
x
1/4
1/2
1
2
4
8
f (x) = 2x
1/4
1/2
1
2
4
8
g(x) = log2 x
-2
-1
0
1
2
3
Reverse coordinates.
y=x
f (x) =
2x
6
5
4
3
f (x) = log2 x
2
-2
-1
-1
-2
2
3 4
5
6
Properties of f(x) = logb x
•Domain = (0, ∞)
•Range = (-∞, +∞)
•X intercept = 1 ; No y-intercept
•Vertical asymptote on y-axis
•Decreasing on 0<b<1; increasing if b>1
•Contains points: (1, 0), (b, 1), (1/b, -1)
•Graph is smooth and continuous
Common Logs and Natural Logs
A logarithm with a base of 10 is a ‘common log’
3
log10 1000 = ______
because 103 = 1000
If a log is written with no base it is assumed to be 10.
log 1000 = log10 1000 = 3
A logarithm with a base of e is a ‘natural log’
0
loge 1 = ______
because e0 = 1
If a log is written as ‘ln’ instead of ‘log’ it is a natural log
ln 1 = loge 1 = 0
Properties & Rules of Logarithms
Basic Properties
Logb b = 1
1 is the exponent to which b must be raised to obtain b. (b1 = b).
Logb 1 = 0
0 is the exponent to which b must be raised to obtain 1. (b0 = 1).
Inverse Properties
logb bx = x
The logarithm with base b of b raised to a power equals that power
b logb x = x
b raised to the logarithm with base b of a number equals that num
For M>0 and N > 0
Product Rule
logb(MN) = logb M + logb N
Quotient Rule
logb M = logb M - logb N
N
Power Rule
p
logb M = p logb M
Logarithmic Property Practice
Quotient Rule
Product Rule
logb(MN) = logb M + logb N
1) log3 (27 • 81) =
logb M = logb M - logb N
N
1) log8 23 =
x
2) log (100x) =
2) Ln
3) Ln (7x) =
Power Rule
p
logb M = p logb M
1) log5 74
2) Log
=
(4x)5 =
3) Ln x2 =
4) Ln x =
e5
11
=
Expanding Logarithms
logb(MN) = logb M + logb N logb M = logb M - logb N
N
p
logb M = p logb M
1) Logb (x2 y )
2)
x
36y4
log6 3
3) log5
x
25y3
4) log2 5x2
3
Condensing Logarithms
logb(MN) = logb M + logb N logb M = logb M - logb N
N
p
logb M = p logb M
Note: Logarithm coefficients
Must be 1 to condense.
(Use power rule 1st)
1) log4 2 + log4 32
4) 2 ln x + ln (x + 1)
2) Log 25 + log 4
5) 2 log (x – 3) – log x
3) Log (7x + 6) – log x
6) ¼ logb x – 2 logb 5 – 10 logb y
The Change-of-Base Property
log a M
log b M =
log a b
Example:
Evaluate log3 7
Most calculators only use:
• Common Log [LOG] (base 10)
• Natural Log [LN] (base e)
It is necessary to use the change
Of base property to convert to
A base the calculator can use.
log10 7
log 3 7 =
log10 3
ln 7
log 3 7 =
ln 3
log3 7 = 1.77