Chapter 5 Study Guide (5.1-5.5)
5.1 Variation Functions
The variable y varies directly as the variable x if
y  kx for some constant k.
k is called the constant of variation.
If y varies directly as x, and y  52 when x  4, find y when x  6.
Step 1
Use y  52 when x  4.
y  kx
52  k · 4
13  k
Step 2
Write the direct
variation equation.
y  kx
y  13x
Step 3
Solve for y when x  6.
y  13x
y  13 · 6
y  78
The variable y varies jointly as the variables x and z if
y  kxz for some constant k.
If y varies jointly as x and z, and y  90 when x  36 and z  5, find y when x  40 and z  3.
Step 1
y  kxz
90  k · 36 · 5
90  180k
0.5  k
Step 2
Write the joint
variation equation.
y  kxz
y  0.5xz
The variable y varies inversely as the variable x if
Step 3
Solve for y when x  40
and z  3.
y  0.5xz
y  0.5 · 40 · 3
y  60
y
k
x for some constant k.
If y varies inversely as x, and y  4 when x  30, find y when x  20.
Step 1
Use y  4 when x  30.
k
y
x
k
4
30
120  k
Step 2
Write the inverse
variation equation.
k
y
x
120
y
x
Step 3
Solve for y when x  20.
120
y
x
120
y
20
y6
Solve each problem.
1. If y varies inversely as x, and y  2 when x  9, find y when x  6. Then
graph the inverse variation function.
Solve each problem.
2. If y varies directly as x, and y  30 when x  20, find y when x  50.
3. If y varies jointly as x and z, and y  150 when x  2.5 and z  12, find y when
5.2. Multiplying and Dividing Rational Expressions
Examples of rational expressions:
3 x 1
x3
,
, and
x x2
2x 2
Undefined at x  0
Undefined at x  0
Undefined at x  2
When simplifying a rational expression:
• Factor the numerator and the denominator completely.
• Divide out any common factors.
• Identify any x-values for which the expression is undefined.
24 x 6
Simplify:
.
x  0, because 8x2 is undefined at x  0.
8x 2
24 x 6 8  3 6 2

x
 3x 4
8
8x 2
Use the Quotient of Powers Property.
x 2  2x  8
.
x2  x  2
Simplify:
First, factor the numerator and the denominator.
x 2  2 x  8 ( x  4)( x  2) ( x  4) ( x  2) ( x  4) x  4




( x  2)( x  1)
( x  2) ( x  1) ( x  1) x  1
x2  x  2
x  2 and x  1
Divide out common factors.
Simplify.
4.
7.
x 2  2x  3
x 2  6x  5
x 3  8x 2
x
x
=______
3
x  4x 2
5
20 x 9
4x 3
6.
8.
x 2  4x
x 2  5x  4
2y
4y 2
2y
x4


= _____
x  4 x  4 x  4 4y 2
5.3 Adding and Subtracting Rational Expressions
Use the least common denominator (LCD) to add rational expressions with different
denominators. The process is the same as adding fractions with different denominators.
Add:
x4
2x

.
x  2x  3 x  1
2
Step 1 Factor denominators completely.
x4
2x
x4
2x



2
x  2 x  3 x  1 ( x  3)( x  1) x  1
Step 2 Find the LCD.
The LCD is the least common multiple of the denominators:
(x  3) (x  1).
Step 3 Write each term of the expression using the LCD.
2x
2x  x  3 
2x 2  6 x


x  1 x  1  x  3  ( x  1)( x  3)
So,
x4
2x
x4
2x 2  6 x



( x  3)( x  1) x  1 ( x  3)( x  1) ( x  1)( x  3)
Step 4 Add the numerators and simplify.
x  4  2x 2  6 x 2x 2  7 x  4

( x  3)( x  1)
( x  3)( x  1)
Add or subtract.
9.
7x  5 4x  1

x 3
x 3
11.
x 1
3x

2
x 4 x2
10.
12.
2x  1 5 x  4

x 1
x 1
4x  1
3

x  3x  2 x  1
2
5.4 Rational Functions
A rational function can be written as a ratio of two polynomials.
a
f x 
k
This is a rational
The graph of this
xh
function.
function is a hyperbola.
There is a vertical asymptote at x  h and the domain is {x | x  h}.
There is a horizontal asymptote at y  k and the range is {y | y  k}.
a
Identify h and k to graph rational functions of the form f  x  
k.
xh
1
Graph g  x  
3.
k  3
x 2
h2
Vertical asymptote at x  2.
Horizontal asymptote at y  3.
The graph of f(x) 
is translated 2
units right and 3 units down.
Identify the asymptotes of each function. Describe the transformation
1
of f  x   . Then graph each function.
x
13. g  x  
1
2
x 1
14. g  x  
1
3
x 1
Vertical asymptote: _________________
Vertical asymptote: _________________
Horizontal asymptote: _______________
Horizontal asymptote: _______________
________________________________________
________________________________________
5.4 Rational Functions (continued)
Use the zeros and the asymptotes of f  x  
px
q x
to graph f(x).
The GCF of p(x) and
q(x) must be 1.
The zeros of f(x) occur where p(x)  0.
The vertical asymptotes of f(x) occur where q(x)  0.
Graph f  x  
x 2  2x  8
.
x 1
Step 1 Find the zeros.
Factor the numerator: x2  2x  8  (x  2) (x  4).
The zeros occur at 2 and 4.
Step 2 Find the vertical asymptotes.
x  1  0 at x  1
x  2  0, so x  2
x  4  0, so x  4
Step 3 Graph.
Plot the zeros at (2, 0) and (4, 0).
Draw the vertical asymptote at x  1.
Make a table of values and plot.
x
y
8
8
0
8
2
8
10
8
Identify the zeros and the vertical asymptotes of each function. Then
graph.
15. f  x  
f x 
x 2  x  12
x2
 x  3  x  4 
x2
16. f  x  
x2  x  6
x 1
_________________________________
Zeros: ___________________________
Zeros: ___________________________
Vertical asymptote: _________________
Vertical asymptote: _________________
5.5 Solving Rational Equations
To solve a rational equation, clear any denominators by multiplying each
term on both sides of the equation by the least common denominator,
LCD.
Solve: x 
12
7.
x
Step 1 The LCD is x. Multiply each term by x.
12
x x 
x  7x
x
This makes the equation a
quadratic equation.
Step 2 Simplify.
x2  12  7x
Step 3 Write in standard form.
x2  7x  12  0
Set one side equal to 0 to
solve a quadratic equation.
Step 4 Factor the quadratic equation.
(x  3) (x  4)  0
Step 5 Set each factor equal to 0.
x30
x40
Step 6 Solve each equation.
x3
x4
x
Check
12
7
x
x3
12
3
347
3
Always check the solutions
to rational equations.
x4
12
4
437
4
Solve each equation.
17.
x
4
 1
2
x
18. x 
6
1
s
19. x  4 
5
x
5.5 Solving Rational Equations and Inequalities (continued)
Check all solutions to rational equations. If the solution to a rational
equation makes the denominator equal to zero, then that solution is NOT a
solution. It is called an extraneous solution.
Solve:
x4 x
10
.
 
x 6 2 x 6
Step 1 The LCD is 2(x  6). Multiply each term by 2(x  6).
x4
x
10
 2  x  6   2  x  6 
 2  x  6
x 6
2
x 6
Step 2 Simplify.
2(x  4)  x(x  6)  10(2)
2x  8  x2  6x  20
Remember to multiply EVERY
term by the LCD.
Step 3 Write in standard form.
x 2  4x  12  0
Step 4 Factor the quadratic equation.
(x  2) (x  6)  0
Step 5 Set each factor equal to 0 and solve.
x20
x60
x  2
x6
x4 x
10
Step 6 Check:
 
x6 2 x6
x  2
x  6 is extraneous.
2  4 2
10
?


2  6 2 2  6
2
10
  1 
This value makes the denominators
8
8
of the original equation equal to 0.
The only solution is x  2.
Solve each equation.
1
x 1 x
20.


x2 x2 5
21.
x x3
4


3 x 1 x 1