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.

• Example: f(x) =

• 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.

3x2 + 2x – 4

x3 - 5x2 + 3

9-1 Rational Functions (cont.)

A point of discontinuity will look like this:

A vertical asymptote is seen in a graph like this:

The rational function f(x) =

has a vertical asymptote at x = 2, illustrated here by the red dashed line.

x2 - 6

x - 2

3x+6

x+2The rational function f(x) =

has a discontinutity at x = -2, illustrated here by the red circle.

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:

a) f(x) =

b) f(x) =

c) f(x) =

X2 + 2x

x + 2

x2 + 7x + 6

x2 + 4x - 12

x - 3

x3 – 6x2 + 9x

2. For each equation, write the equation for the vertical and horizontal asymptotes.

a) f(x) =

b) f(x) =

c) f(x) =

1

x - 3

3

(x + 1)(X – 5)

2

x2 – 2x - 8

9-1 Examples (cont.)

1. Solutions:

a) f(x) =

b) f(x) = =

c) f(x) = =

X2 + 2x

X + 2

Only x = -2 causes the denominator to be 0.

a.

In the graph, we see a discontinuity at x = -2

X2 + 7x + 6

X2 + 4x - 12

(X + 6)(x + 1)

(X + 6)(x – 2)

x= 3, x = -3, and x=0 all cause the denominator to be 0.

In this graph, we see a discontinuity at x = -6, and a vertical asymptote at x=2

b.

X - 3

X3 + 9x

X - 3

X(x-3)(x+3)

Here, both x = -6 and x = 2 cause the denominator to be 0.

c.

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.

a) f(x) =

b) f(x) =

c) f(x) =

1

X - 33

(X + 1)(X – 5)

2

X2 – 2x - 8

a.

Notice that the denominator = 0 when x = 3.

Notice that the denominator = 0 when x = -2 or 4.

Notice that the denominator = 0 when x = -1 or 5.

From the graph, the vertical asymptote is x=3. There is also a horizontal

asymptote along y=0.

b.

From the graph, the vertical asymptotes are x= -2 and

x=4. There is also a horizontal asymptote along y=0.

b. c.

From the graph, the vertical asymptotes are x= -1 and

x=5. There is also a horizontal asymptote along y=0.

9-1 Problems

1. Find f(-2) and f(4) for f(x)=2. Determine the points of discontinuity and the vertical asymptotes,

first by determining the x values that cause the denominator to equal zero, and then by graphing, for f(x) =

3. Give the equations for the horizontal and vertical asymptotes of f(x) =

1)f(2) = undefined (vertical asymptote) 2)x = -1, x= -4 3) x = -10 (V), x = 1 (v), y = 0 (H)

f(4) = 1x = -4 is vertical asymptote

x = -1 is a discontinuity

X + 2

X2 - x - 6

x2 + x

X2 + 5X + 4

2

X2 + 9X -10

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. This means that a direct relationship occurs if y = k * x, where k is any

constant, other than zero. 2. For example: y = 4.5*x means y varies directly as x, and the constant of

variation (k) is equal to 4.5.

2. We say that x and y are inversely related if y is a multiple of 1/ x1. This means that an inverse relationship occurs if y = k / x, where k is

any constant other than zero.2. For example: y = 2 / x means y varies inversely as x, with constant of

variation 2.3. Another way of expressing inverse relationships is: x*y = k

3. We say that x and y vary jointly if y is a multiple of two or more variables.

1. 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.

2. 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. Since y varies directly as x, our equation is y = k * k.

2. Start by using what you know. We know when x = 1.5, y = 3.

3. Plugging this in to y = k*x gives 3 = k * 1.5.

4. Solving for k gives k = 3 / 1.5, or k = 2.

5. Now, set up your second equation, using y = 2 * x.

6. 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

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.

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.

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

1. Use x = 4, y = -12, and z = 5 to solve the following:A. Y varies directly as x, what is y when x = 3?B. Y varies jointly as x and z, what is y when x = 9 and z = 2

2. For the following, state if the relationship is direct, inverse, or join variation, and find the constant of variation, k:

A. X*Y = 4 C. X * Z = -2 * YB. 12*Y = X D. X = 3 / Y

3. 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.

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:

• Just like with a fraction, we can eliminate x/x and (x-2)/(x-2) because anything divided by itself is equal to 1.

x*(x-2)(x+3)x*x*(x-2)=

x*(x-2)(x+3)x2(x-2)

= (x+3)x

=x*(x-2)(x+3)x*x*(x-2)

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:

• For any rational function, we divide:

A

B =

C

D *

AC

BD As long as B ≠ 0 and D ≠ 0. For example: 3x

x2 =

(x+1)

(x+2)*3x(x+1)

X2(x+2)

A

B =

C

D ÷

AD

BC As long as B ≠ 0, C ≠ 0 D ≠ 0. For example:

4x

x3=

(x+7)

(x-2)*4x(x-2)

X3(x+7)

9-3 Example

1. Give the simplified form of each rational equation: x2

x+5

3x

x* =

3x*x2

x(x+5)=

3x3

x(x+5)

3x3 2

x(x+5)= =

3x2

x+5

x

x-2

x2-4

x* =

x*(x2-4)

x*(x-2)=

x2-4

(x-2)= =

x*(x2-4)

x*(x-2)

(x+2)(x-2)

(x-2)

(x+2)(x-2)

(x-2)= = x+2

x+15

6x

4x4

2x÷ =

2x*(x+15)

6x*4x4=

2x*(x+15)

24x5

2x*(x+15)

24x5 4= =

x+15

12x4

Note:

2/24 = 1/12

x/x5 = 1/x4

(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

9-3 Problems

1. Multiply:

2. Multiply and simplify:

3. Divide:

4. Divide and simplify:

7x

x+5

3x(x+2)

x-10*

x2 (y+2)

x7

y5

4y2(y+2)*

xy3

y-8

3x2

y+4

y(x+4)

y4(y-5)

x2(x+4)

x4÷

÷

21x2(x+2)

(x-10)(x+5)

1.y3(y+2)

4x5

2.xy3(y+4)

3x2 (y-8)3.

x2

y3(y-5)

4.

9-4 Adding and Subtracting Rational Functions

• Adding and subtracting rational functions is a lot like adding and 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.

• For example:

• 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.

2.

1

3

3

8+ =

1*8

3*8

3*3

3*8+ =

8

24

9

24+ =

17

24

x+2

4x2

3

7x+ =

7(x+2)

7*4x2

4x*3

4x*7x+ =

7x+14

28x2

12x

28x2+ =

19x+14

28x2

2a

7b2

3abc

11ac+ =

2a*11ac

7b2*11ac

3abc*7b2

11ac*7b2+ =

22a2c

77ab2c+

21ab3c

77ab2c=

22a2c + 21ab3c

77ab2c

9-4 Examples

1. Perform the addition or subtraction of the following rational equations.

3xy

2(y+1)

y

15x2

Since these denominators have nothing in common, we multiply the numerator and denominator of each fraction by the opposite denominator.

3xy*15x2

2(y+1)*15x2

y*2(y+1)

15x2*2(y+1)Now we have a common denominator, we can simplify, then subtract.

45x3y

30x2(y+1)

2y(y+1)

30x2(y+1)

45x3y – 2y(y+1)

30x2(y+1)

Sometimes only one portion of the rational equations needs adjusting.

3x

2y

x2

4y3

3x*2y2

2y*2y2

x2

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.

6xy2

4y3

x2

4y3

6xy2 – x2

4y3

9-4 Examples

3. It is possible to combine the concepts of rational multiplication and division with the concepts of rational addition and subtraction.

1

2y

y

x2

5

+

4

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.

1*x2

2y*x2

y*2y

x2 *2y

5*6y

1*6y

+

4

6y

The LCD for the numerator is 2x2y and the LCD for the denominator is 6y. Now, combine the numerator and denominator equations.

1*x2

2y*x2

y*2y

x2 *2y

5*6y

1*6y

+

4

6y

2x2+2y2

2x2y

30y - 4

6y

Once the numerator rational and denominator rational are taken care of, the equation is a division of two rationals, just like section 9-3.

2x2+2y2

2x2yx

6y

30y - 4

9-4 Problems

1. Find the Least Common Denominator

2. Solve the following equations:

3. Solve the following, using sections 9-3 and 9-4 material.

x

7y3x2

5y

4x

+

3y

2x2

4x2(x+7x2y3)

7y3(5xy-6y)

y

4x2y

y

3xy3

2x

3x(y+2)

y2

3xy+6x

x + y

10x2

4x2

5xy

12x2y3 , 3xy+6x , 10x2yx2

19x2

y

4(x+3)

x+4

72xy

y

x (x+4)-72y2

72xy

4(x+3)x2+19x2y

76x2(x+3)

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 helpful in solving inequalities.

• 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 changesFor example:

-3x < 5

x > - 5/3

9-5 Examples

1. Solve the following rational equation for x.

3

2(x+1)

1

15x

The LCD is 30x(x+1). Multiply the LCD by both side of the equation.

Next, multiply the 30x and 30(x+1) into the rational expressions.

3

2(x+1)

1

15x30x(x+1) 30x(x+1)* *

3

2

1

1530x 30(x+1)* *

90x

2

30(x+1)

15

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.

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+1Start by subtracting the right side to the left.

Next, simplify. First, get a common denominator. The LCD is 7(x+2)(x+1)

Combine the two parts now that there is a common denominator.

<x

7(x+2)

x

x+1< 0

x(x+1)

7(x+2)(x+1)

x*7(x+2)

7(x+2)(x+1)<

0

x(x+1) – x(x+2)

7(x+2)(x+1)< 0

x2+x – x2 – 2x

7(x+2)(x+1)< 0

-x

7(x+2)(x+1)<0

x

(x+2)(x+1)> 0*-7

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.

Multiply each side by -7 to remove these constants. Don’t forget to change inequality.

x

(x+2)(x+1)> 0

-2 -1 0

Plugging x=-3 into the inequality gives -1.5 > 0 which is false.

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☺ ☺

Thus, the final solution is -2 < x < -1 and x > 0.

9-5 Problems

1. Solve the following rational equations:

2. Solve the following rational inequalities:

z -1

z2

z-1

z3

x

x+2

1

2

3

5x+5

1

x+1== = +

2

5x = 2 z = 1 x = -2

1

x+1

x

2>

x-2

x

4

x+4<

x < -2 and -1<x<1 -4 <x< -2 and 0<x<4