### Chapter 9

```Chapter 9
Exploring Rational Functions
Dan Box
9-1 Rational Functions
• A function, f(x) is a rational function if it is the
division of two polynomial functions, meaning it
can be written:
• f(x) = p(x) / q(x) , where p(x) and q(x) are polynomial
functions, and q(x) ≠ 0.
3x2 + 2x – 4
• Example:
f(x) =
x3 - 5x2 + 3
•
It is important that q(x), the denominator, is not 0.
• If a value of x makes the denominator equal to 0, that x value will
either be a vertical asymptote or a point of discontinuity.
• A point of discontinuity is appears as a whole in the graph at the
x value.
• A vertical asymptote is an x value that the graph comes very close to, but
never actually touches.
9-1 Rational Functions (cont.)
A point of discontinuity will look like this:
The rational function f(x) =
3x+6
x+2
has a discontinutity at x = -2, illustrated
here by the red circle.
A vertical asymptote is seen in a graph like this:
x2 - 6
The rational function f(x) =
x-2
has a vertical asymptote at x = 2,
illustrated here by the red dashed line.
9-1 Examples
1.
For each equation, find the x values that result in vertical asymptotes or
points of discontinuity. Then, use a calculator to check what each x is:
X2 + 2x
a)
f(x) =
x+2
x2 + 7x + 6
b)
f(x) =
c)
f(x) =
2.
x2 + 4x - 12
x-3
x3 – 6x2 + 9x
For each equation, write the equation for the vertical and horizontal
asymptotes.
a)
f(x) =
1
x-3
b)
f(x) =
3
(x + 1)(X – 5)
c)
f(x) =
2
x2 – 2x - 8
9-1 Examples (cont.)
1.
Solutions:
a)
f(x) =
b)
f(x) =
c)
f(x) =
X2 + 2x
X+2
X2 + 7x + 6
X2
+ 4x - 12
X-3
X3
a.
+ 9x
b.
In the graph, we see a
discontinuity at x = -2
Only x = -2 causes the
denominator to be 0.
=
=
(X + 6)(x + 1)
(X + 6)(x – 2)
X-3
X(x-3)(x+3)
Here, both x = -6 and x = 2
cause the denominator to be 0.
x= 3, x = -3, and x=0 all cause
the denominator to be 0.
c.
In this graph, we see a discontinuity at
x = -6, and a vertical asymptote at x=2
In this graph, we see a
discontinuity at x = 3, and vertical
asymptotes at x=0 and x = -3
9-1 Examples (cont.)
2. For each equation, write the equation for the vertical and horizontal
asymptotes.
Notice that the denominator = 0 when x = 3.
1
a)
f(x) =
b)
f(x) =
c)
f(x) =
X-3
3
Notice that the denominator = 0 when x = -1 or 5.
(X + 1)(X – 5)
2
Notice that the denominator = 0 when x = -2 or 4.
X2 – 2x - 8
a.
From the graph, the
vertical asymptote is x=3.
There is also a horizontal
asymptote along y=0.
b. b.
From the graph, the vertical
asymptotes are x= -1 and
x=5. There is also a horizontal
asymptote along y=0.
c.
From the graph, the vertical
asymptotes are x= -2 and
x=4. There is also a horizontal
asymptote along y=0.
9-1 Problems
X+2
1.
2.
Find f(-2) and f(4) for
f(x)=
X2 - x - 6
Determine the points of discontinuity and the vertical asymptotes,
first by determining the x values that cause the denominator to
x2 + x
equal zero, and then by graphing, for f(x) =
X2 + 5X + 4
3.
Give the equations for the horizontal and vertical asymptotes of
f(x) =
2
X2 + 9X -10
x = -1 is a discontinuity
f(4) = 1
1)
f(2) = undefined (vertical asymptote)
x = -4 is vertical asymptote
2)x = -1, x= -4
3) x = -10 (V), x = 1 (v), y = 0 (H)
9-2 Direct, Inverse, and Joint Variation
•
There are several ways that x and y can be related in a
function.
1.
We say that x and y are directly related if y is a multiple of x.
1.
2.
2.
We say that x and y are inversely related if y is a multiple of 1/ x
1.
2.
3.
3.
This means that a direct relationship occurs if y = k * x, where k is any
constant, other than zero.
For example: y = 4.5*x means y varies directly as x, and the constant of
variation (k) is equal to 4.5.
This means that an inverse relationship occurs if y = k / x, where k is
any constant other than zero.
For example: y = 2 / x means y varies inversely as x, with constant of
variation 2.
Another way of expressing inverse relationships is: x*y = k
We say that x and y vary jointly if y is a multiple of two or more
variables.
1.
2.
This means that joint variation occurs if we say y = k * x * z , where x
and z are variables that are not zero, and k is a constant that is not
zero.
For example, we know Area = base * height on a rectangle. This means
A = 1 * B * H. Thus we say the area of a rectangle varies jointly with the
width and the height.
9-2 Examples
1.
Direct variation: Given y varies directly as x, and when
x = 1.5, y = 3 , find y when x = 4.
1.
2.
3.
4.
5.
6.
Since y varies directly as x, our equation is y = k * k.
Start by using what you know. We know when x = 1.5, y = 3.
Plugging this in to y = k*x gives 3 = k * 1.5.
Solving for k gives k = 3 / 1.5, or k = 2.
Now, set up your second equation, using y = 2 * x.
This gives y = 2 * 4, which means when x = 4, y = 8.
Note: graphs of direct relationships are straight lines through
the origin (0,0). Here is the graph of y = 2x, from the example.
9-2 Examples
1. Inverse Relationships: If y varies inversely as x, and when
x = 5, y = 8, what is y when x = -2 ?
1. Remember that in inverse relationships, y = k / x.
2. Use the given information: 8 = k / 5.
3. Cross multiplying gives: 8 * 5 = k * 1, or 40 = k.
4. State the equation: y = 40 / x.
5. Solve using the new equation. y = 40 / -2. This gives y = -20, the solution.
6. Alternatively, you could use the initial equation xy = k.
7. This still gives 8 * 5 = k, k = 40. It would also give -2*y = 40, with the same
result : y = -20.
Note: Graphs of inverse relationships are curved lines, with
vertical and horizontal asymptotes at x = 0, y = 0. The graph of
the solution function y = 40 / x is given above.
9-2 Examples
1. Varying Jointly: If y varies jointly as x and z, and when x = 2 and z = 4 , y = 12, what is y when x = 5 and z = 2?
1. Remember that in joint variation, y = k*x*z.
2. Use the given information: 12 = k*(-2)*(4).
3. Solving for k gives: 12 = k*(-8) , 12/-8 = k , k = -1.5
4. State the equation: y = (-1.5)*x*z .
5. Solve using the new equation. y = (-1.5)*(5)*(2). This gives y = -15, the
solution.
9-2 Problems
Use x = 4, y = -12, and z = 5 to solve the following:
A.
B.
2.
For the following, state if the relationship is direct, inverse,
or join variation, and find the constant of variation, k:
A.
B.
3.
Y varies directly as x, what is y when x = 3?
Y varies jointly as x and z, what is y when x = 9 and z = 2
X*Y = 4
12*Y = X
C. X * Z = -2 * Y
D. X = 3 / Y
Assume suburb population varies jointly with distance from
major city and the number of train stations available to the
city. The distance from Naperville to Chicago is 27 miles,
and Naperville has 2 train stations. The population of
Naperville is 129,600. Given this, how many train stations
would you expect in Schaumburg, IL, if Schaumburg has a
population of 57,600 and is 24 miles from Chicago?
1A)y=-9 1B)y= -10.8 2A) Inverse, k = 4 2B) direct, k = 1/12 2C) joint, k = -1/3
2D) inverse, k = 3 3) k = 2,400 Schaumburg would have 1 train station.
1.
9-3 Multiplying and Dividing Rational
Expressions
• Often the first step in multiplying or dividing
rational expressions involves simplifying.
• Just like with a rational number, like 9/15, you
can simplify by removing a common factor from
the top and bottom of the expression: 9/15 =
(3*3)/(3*5) = 3 / 5
• In a rational expression, we look to remove
variables as well numbers.
For example: x*(x-2)(x+3) x*(x-2)(x+3) = x*(x-2)(x+3) = (x+3)
x2(x-2)
=
x*x*(x-2)
x*x*(x-2)
x
• Just like with a fraction, we can eliminate x/x and
(x-2)/(x-2) because anything divided by itself is
equal to 1.
9-3 Multiplying and Dividing
Rational Expressions
• Remember: When multiplying two fractions, you
multiply the numerators and multiply the
denominators. When dividing, multiply the
denominator of the first rational with the
denominator of the second, and the denominator
of the first with the numerator of the second.
• For any rational function, we multiply:
A
B
*
C
D
=
AC
BD
As long as B ≠ 0 and D ≠ 0. For example:
3x (x+1) 3x(x+1)
=
x2 * (x+2) X2(x+2)
• For any rational function, we divide:
A
B
÷
C
D
=
BC
As long as B ≠ 0, C ≠ 0 D ≠ 0. For example:
4x (x+7) 4x(x-2)
=
x3 * (x-2) X3(x+7)
9-3 Example
1.
Give the simplified form of each rational equation:
x2
*
3x
=
x+5
x
x
x2-4
*
x-2
3x*x2
x(x+5)
x
=
3x3
=
x(x+5)
3x3 2
x(x+5)
=
3x2
x+5
(x+2)(x-2) (x+2)(x-2)
x*(x2-4) x*(x2-4)
x2-4
=
=
=
=
=
= x+2
(x-2)
(x-2)
x*(x-2) x*(x-2)
(x-2)
x+15
x+15 4x4
2x*(x+15) 2x*(x+15) 2x*(x+15)
÷
=
=
=
=
4
5
5
4
12x4
6x
2x
6x*4x
24x
24x
(x+4)(x-3)
-3x4
÷
x-3
2x8
=
2x8(x+4)(x-3)
-3x4(x-3)
=
2x8(x+4)(x-3)
-3x4(x-3)
=
2x4(x+4)
-3
Note:
2/24 = 1/12
x/x5 = 1/x4
9-3 Problems
3x(x+2)
*
x+5 x-10
x7
xy3
Divide:
y-8
3x2
y+4
y(x+4) x2(x+4)
÷
y4(y-5)
x4
2.
y3(y+2)
3
3. xy (y+4)
Divide and simplify:
÷
4y2(y+2)
3x2 (y-8)
y3(y-5)
4.
x2
4.
*
21x2(x+2)
3.
y5
1.
Multiply and simplify:
x2 (y+2)
4x5
2.
7x
Multiply:
(x-10)(x+5)
1.
Rational Functions
•
•
•
subtracting fractions.
Just like adding or subtracting a fraction, the first step is to form a
common denominator between all the functions involved. To do this,
find the Least Common Multiple of the denominators, and then add
the numerators.
1
3
1*8 3*3
8
9
17
For example:
+
=
+
=
+
=
3
•
•
8
3*8
3*8
24
24
24
You can use this same process for rational functions. The goal is to
get the same function as the denominator of both rational equations.
For example:
1.
x+2
3 7(x+2) 4x*3 7x+14 12x
19x+14
+
+
+
=
=
=
4x2 7x 7*4x2 4x*7x
28x2 28x2
28x2
3. 2a
7b2
+
3abc 2a*11ac 3abc*7b2
=
11ac
+
7b2*11ac
11ac*7b2
=
22a2c
77ab2c
+
21ab3c
77ab2c
22a2c + 21ab3c
=
77ab2c
9-4 Examples
1. Perform the addition or subtraction of the
following rational equations.
3xy
y
3xy*15x2
y*2(y+1)
2(y+1)
15x2
2(y+1)*15x2
15x2*2(y+1)
Since these denominators have
nothing in common, we multiply
the numerator and denominator
of each fraction by the opposite
denominator.
45x3y
Now we have a common
denominator, we can simplify, then
subtract.
2y(y+1)
30x2(y+1) 30x2(y+1)
45x3y – 2y(y+1)
30x2(y+1)
Sometimes only one portion of the rational equations needs adjusting.
3x
x2
3x*2y2
x2
6xy2
x2
6xy2 – x2
2y
4y3
2y*2y2
4y3
4y3
4y3
4y3
Here, the least common denominator is actually 4y3. We can turn 2y into 4y3 by
multiplying the top and bottom of the first rational fraction by 2y2.
9-4 Examples
3.
It is possible to combine the concepts of rational
multiplication and division with the concepts of rational
1
y
1*x2
x2
2y*x2
2y
5
+
+
y*2y
1*x2
x2 *2y
2y*x2
+
y*2y
2x2+2y2
x2 *2y
2x2y
2x2+2y2
2x2y
4
5*6y
4
5*6y
4
30y - 4
6y
1*6y
6y
1*6y
6y
6y
The numerator of this
equation is the sum of
two rational functions,
and the denominator is
the different of two
rational equations. Start
by finding the common
denominators for the top
and bottom separately.
The LCD for the
numerator is 2x2y and
the LCD for the
denominator is 6y. Now,
combine the numerator
and denominator
equations.
Once the numerator
rational and
denominator rational
are taken care of, the
equation is a division
of two rationals, just
like section 9-3.
x
6y
30y - 4
9-4 Problems
1.
Find the Least Common Denominator
y
y
2x
y2
x+y
4x2
4x2y
3xy3
3x(y+2)
3xy+6x
10x2
5xy
12x2y3 , 3xy+6x , 10x2y
Solve the following equations:
19x2
4(x+3)
x+4
y
72xy
x
4(x+3)x2+19x2y
3.
Solve the following, using sections 9-3 and 9-4 material.
x
7y3
+ x2
5y
3y
7y3(5xy-6y)
4x
2x2
(x+4)-72y2
y
76x2(x+3)
x2
72xy
2.
4x2(x+7x2y3)
9-5 Solving Rational Equations
• Any equation that involves one or more rational
expressions is a rational equation
• Rational equations are usually solved by
multiplying each side of the equation by the least
common denominator (LCD).
• It is important to check solutions to rational
equations to be sure that you have no multiplied
by zero along the way. The best way to check
this is to plug your solutions back in to the initial
equation and make sure that you get a real
solution.
9-5 Solving Rational Inequalities
• Solving rational inequalities is much like solving
for rational equations, but sets one rational
equation as greater or less than another.
• Solving so that one side is equal to zero is
• It may also be helpful to use a number line to
test possible solutions.
• Remember, when working with inequalities, if
you multiply or divide by a negative number, the
direction of the inequality changes
For example:
-3x < 5
x > - 5/3
9-5 Examples
1. Solve the following rational equation for x.
3
2(x+1)
1
15x
3
30x(x+1) *
2(x+1)
The LCD is 30x(x+1).
Multiply the LCD by
both side of the
equation.
90x
30(x+1)
2
15
1
15x
* 30x(x+1)
Here we see some common terms
between the numerator and
denominator that can be canceled out.
Simplify the fractions, and distribute
the expression in the numerator in
the right equation.
45x
2(x+1)
Now solve using algebra.
30x *
3
1
2
15
* 30(x+1)
Next, multiply the 30x and 30(x+1)
into the rational expressions.
45x
2x+2
45x-2x-2
0
43x-2
0
43x
2
x
2
43
Plugging the solution x=2/43
into the equation, we get the
left side is equal to about
1.433, and the right side is
equal to 1.433. Since each
side is equal our result is true.
9-5 Examples
2. Solve the following rational inequality for x
x
7(x+2)
<
5
x
x
x+1
7(x+2)
x+1
Start by subtracting the
right side to the left.
7(x+2)(x+1)
7(x+2)(x+1)
< 0
x
> 0
From here, we know that x is zero
or undefined at x=0, -2, and -1.
Using this information, use a
number line to test ranges.
-x
< 0
7(x+2)(x+1)
<
0
Next, simplify.
x2+x – x2 – 2x
Combine the two parts now that
there is a common denominator.
(x+2)(x+1)
x*7(x+2)
First, get a common denominator.
The LCD is 7(x+2)(x+1)
x(x+1) – x(x+2)
7(x+2)(x+1)
< 0
x(x+1)
7(x+2)(x+1)
Plugging x=-3 into the inequality gives
-1.5 > 0 which is false.
x
<0
(x+2)(x+1)
> 0*-7
Multiply each side by -7 to remove
these constants. Don’t forget to
change inequality.
X= -1.5 gives 6 > 0 which is true
X = -0.5 gives -0.66 > 0 which is false.
X = 1 gives .166 > 0 which is true.
X
☺ X ☺
-2
-1
0
Thus, the final solution
is -2 < x < -1 and x > 0.
9-5 Problems
1. Solve the following rational equations:
z -1
= 2
z2
z-1
= z3
z=1
x+2
1
3
1
2
5x+5
= x+1 + 5
x = -2
x
x=2
2. Solve the following rational inequalities:
> 2
x < -2 and -1<x<1
x+1
x
x-2
x
4
< x+4
-4 <x< -2 and 0<x<4
1
```