Rational Equation
& Inequality Review
Reciprocals
Given a function
The reciprocal
function is
f (x )
1
f ( x)
Some Properties
f ( x)
1
f ( x)
is less than -1
is between -1 and 0
-1
-1
is between -1 and 0
is less than -1
0
a vertical asymptote
is between 0 and 1
is greater than 1
1
1
is greater than 1
is between 0 and 1
More Properties
f ( x)
1
f ( x)
Positive
Positive
Negative
Negative
Increasing
Decreasing
Decreasing
Increasing
Reciprocal Function
y  x  6x  5
2
Positive Interval
Negative Interval
Increasing Interval
Decreasing Interval
X-intercept
Y-intercept
Local Max
Local Min
Vertical Asymptote
Horizontal Asymptote
x  1 or x  5
1 x  5
x3
x3
x  1, 5
y 5
N/A
 3,  4 
N/A
N/A
y
1
x2  6 x  5
x  1 or x  5
1 x  5
x  3, x  1
x  3, x  5
none
y 1
5
3, 1
4


N/A
x  1, 5
y0
Quadratic Example
IIV
II
III
Original:
0 <><1,
y-1
-1,
yy<positive,
<1,0,
increasing
positive,
negative,
Reciprocal:
increasing
0 < y < 1,
Reciprocal:
ypositive,
-1
<><1,
-1,
y positive,
< 0,
decreasing
negative,
decreasing
y  x2  4
I
II
III
IV
1
y 2
x 4
X-intercept
Points of intersection
of original
between
functionthe
becomes
original
asymptote
function and
of the
its
reciprocal
reciprocal
function
Example
Use your knowledge of rational functions and the transformations
1
of a square root functions to sketch y 
.
2 x  1
Rational Function
Vertical asymptote(s):
zero(s) of the denominator that’s
not a zero for the numerator
Horizontal Asymptote: based on degrees of the
numerator and the denominator
x-intercept(s):
zero(s) of the numerator
y-intercept:
value of the function when x = 0
Example
Find (if exists) the x, y-intercepts, and all asymptotes of
y
x2  x  6
 x  1
2
 x  3 x  2 

2
 x  1
X-intercept: zero(s) of numerator
V.A. is x  1.
x  3, 2
Y-intercept: x  0
y  6
H.A. is y  1.
Example
Find (if exists) the x, y-intercepts, and all asymptotes of
x 2  x  2 ( x  2)( x  1) x  2

y

2
x 1
x 1
( x  1)( x  1)
X-intercept: zero(s) of numerator
x  2 1is no longer an x-int.
V.A. is x  1.
Y-intercept: x  0
H.A. is y  1.
y2
x  1is no longer a V.A.
Example
Write an equation for a rational function whose graph has all of
the indicated features:
• a vertical asymptote with equation x = -2
• a horizontal asymptote with equation y = -1
• x -intercept at x = 3
 V . A. x  2 
 x-int at x  3   H . A. at y  1 
denominator ( x  2) numerator ( x  3) ratio of leading coeff. is  1
1
y
( x  2)
( x  3)
y
( x  2)
( x  3)
y
( x  2)
Example
Create a rational function that has the graph with the
given features:
only one vertical asymptote at x = -1,
horizontal asymptote at y =-3,
and x-intercepts of -2 and 4.
X  intercept is x  -2, x  4
H.A. is y  3 :
numerator is
Same degree in numerator
( x  2)  ( x  4)
and denominator
Only 1 V.A. but degree is 2
a3 (x( x2)2) (x( x4)4)
y
((xx1)1)22
denominator is
( x  1) 2
Find the Equation
V.A. is x  1:
denominator is ( x  1)
H.A. is y  0 :
a
y
x 1
numerator has lower degree
than denominator.
y-intercept =  0.5:
a
0.5 
, a  0.5
0 1
0.5
y
x 1
 0, 0.5
Find the Equation
xa
Given y 
and the following conditions :
bx  c
V . A. x  2, x  int  3, y  int  2.
Find the values of a, b, and c.
V.A. is x  2:
y  int is  2 :
2 is zero of denominator
a
a 3
2  , c 

c
2 2
2b  c  0
x  int is 3:
3 is zero of numerator.
3  a  0, a  3
c 3
b

2
4
Example
x2
For the rational function y  2
determine the following
x  x6
but do not graph.
• State any asymptotes. V . A. : x  3, H . A. : y  0
• State the x and y-intercepts. x-int : none, y -int : 1
3
• State the coordinates of any holes. 2, 1
5

• State the domain and range.

D   x x  , x  2, 3

R  y y  , y  0, 1
x2
x2
1
y 2


, x  2
x  x  6  x  2   x  3 x  3
5

Example
The graphs of six rational functions are given. Six equations are
listed below. Match each graph with its equation.
(b)
(c )
(a)
No V.A., with a hole
0.5
1. y  2
x 1
Two V.A.'s
2. y 
x
 x  1 x  2 
No V.A.
x2 1
3. y 
x 1
Example
The graphs of six rational functions are given. Six equations are
listed below. Match each graph with its equation.
V.A. x  1
1. y 
1
 x  1
2
V.A. x  1
Oblique
1
2. y 
x 1
x2
3. y 
x 1
Main Steps
1. Find and plot all the asymptotes.
2. Find and plot all intercepts if they exist.
3. Use test values to check the function’s
behavior about the asymptotes.
4. Connect the curves.
x 2  4 x  3  x  3 x  1 x  1
y 2


, x  3
x  x  6  x  3 x  2  x  2
4 3
V.A.:
x2
H.A.:
y 1
2
1
X-int.:  1
Y-int.:  0.5
Hole : 3, 2 



5
1
x  10000, y  1.0003
2
x  10000, y  0.9997
3
x  2.01, y  301
4
x  1.99, y  299
x  1 x  1
x2  1


y 2
x  4 x  3
4 x  3x
V.A.:
H.A.:
X-int.:
6 5
3
x  0,
4
1
y
4
4 3
Cross over:
Point where the graph crosses the
horizontal/oblique asymptote
2
1
1,1
Set y  H . A.
Y-int.:
1
2
3
4
x2  1
1

4 x 2  3x 4
none
x  1000, y  0.2498
x  1000, y  0.2502
x  0.01, y  33
x  0.01, y  34
3 x  4, x 
5
6
x  0.74, y  15
x  0.76, y  14
4
3
Example
2
x
Determine the equation(s) of all asymptote(s) of y  2  4 .
x 1
Sketch the resulting function . (Be sure to label all asymptotes
with their equation(s) and clearly show all intercepts)
x2  4  x  2   x  2
y 2

x 1
x2  1
V.A.: none
H.A.: y  1
x-int: 2, 2
y -int:  4
Solving Rational Inequality
1. Rearrange the inequality with 0 on one side.
2. Common denominator all terms and write as
a single fraction.
3. Identify all the roots by completely factoring
the numerator and denominator.
4. The roots will divide the domain into sections
where the value of the rational will have the
same sign within each section.
5. Use test values within each region to
determine the validity of the inequality.
Example
x  3 x 1

Determine the solution set for the inequality
.
x  2 x 1
Move everything to one side:
x  3 x 1

0
x  2 x 1
Common denominator:
 x  3   x  1   x  1   x  2 
 0,
 x  2    x  1
x 2  2 x  3  x 2  3x  2
 0,
 x  2    x  1
x 5
0
 x  2    x  1
The zeros are  1,2,5
Example
x  3 x 1

Determine the solution set for the inequality
.
x  2 x 1
The zeros are  1,2,5
x  1 1  x  2 2  x  5 5  x
( x  1)




( x  2)




( x  5)




f




The solution is  x x  , x  1 or 2  x  5
Example
Solve for x.
4
3

2
x 1 x  2
  x  1 x  2 
4  x  2   3  x  1  2  x  2  x  1
x  11  2  x  1 x  2 
0  2  x 2  x  2   x  11,
0  2 x  x  15
2
0   2 x  5  x  3
5
x  or  3
2
Example
A box with a square base and a volume of 100 cm3 is to be built.
The material for the top and bottom costs $2 per 100 cm2 and the
material for the sides costs $3 per 100 cm2. Use the diagram, express
the cost of building the box as a simplified rational function of x.
x 2 y  100
Cost  CostTop  Bottom  Cost Sides
100
y 2
x
2
2
4
x
CostTop  Bottom  2 x 2 

100 100
3 12 x 100 12
CostSides  4 xy 

 2 
100 100 x
x
4 x 2 12 x3  300
 Cost 
 
100 x
25 x
Example
Pipe P takes 10hrs less than smaller pipe Q when they fill a tank
separately. Together, it takes 12hrs to fill up the tank. Find the
time taken by pipe Q to fill the tank alone.
Let x represent the time it takes for pipe Q to fill the tank.
x - 10 represent the time it takes for pipe P to fill the tank.
Pipe P Pipe Q
Time takes
to fill tank
Fraction filled
per hour
x  10
x
1
x  10
1
x
1
1
1


x x  10 12
1
x  x  10  ( x)( x  10)
12
24 x  120  x 2  10 x
x 2  34 x  120  0, x  30, x  4
The time it takes for pipe Q to fill the tank is 30 hrs.