Chapter 9 Exploring Rational Functions

1
Chapter 9Exploring Rational Functions
Dan Box
2
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.

3
9-1 Rational Functions (cont.)

• A point of discontinuity will look like this

3x6 x2
The rational function f(x) has a
discontinutity at x -2, illustrated here by the
red circle.
A vertical asymptote is seen in a graph like this
4
9-1 Examples

• 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
• f(x)
• f(x)
• f(x)

• For each equation, write the equation for the
  vertical and horizontal asymptotes.
• f(x)
• f(x)
• f(x)

5
9-1 Examples (cont.)

• Solutions
• f(x)
• f(x)
• f(x)

Only x -2 causes the denominator to be 0.
Here, both x -6 and x 2 cause the denominator
to be 0.
x 3, x -3, and x0 all cause the denominator
to be 0.
a.
b.
c.
In the graph, we see a discontinuity at x -2
In this graph, we see a discontinuity at x 3,
and vertical asymptotes at x0 and x -3
In this graph, we see a discontinuity at x -6,
and a vertical asymptote at x2
6
9-1 Examples (cont.)

• 2. For each equation, write the equation for the
  vertical and horizontal asymptotes.
• f(x)
• f(x)
• f(x)

Notice that the denominator 0 when x 3.
Notice that the denominator 0 when x -1 or 5.
Notice that the denominator 0 when x -2 or 4.
a.
b.
b.
c.
From the graph, the vertical asymptote is x3.
There is also a horizontal asymptote along y0.
From the graph, the vertical asymptotes are x -2
and x4. There is also a horizontal asymptote
along y0.
From the graph, the vertical asymptotes are x -1
and x5. There is also a horizontal asymptote
along y0.
7
9-1 Problems
X 2 X2 - x - 6

• Find f(-2) and f(4) for f(x)
• 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)

x2 x X2 5X 4
2 X2 9X -10

• f(2) undefined (vertical asymptote)
  2)x -1, x -4 3) x -10 (V), x 1 (v), y 0
  (H)
• f(4) 1 x -4 is vertical asymptote
• x -1 is a discontinuity

8
9-2 Direct, Inverse, and Joint Variation

• There are several ways that x and y can be
  related in a function.
• We say that x and y are directly related if y is
  a multiple of x.
• This means that a direct relationship occurs if y
  k x, where k is any constant, other than
  zero.
• For example y 4.5x means y varies directly
  as x, and the constant of variation (k) is equal
  to 4.5.
• We say that x and y are inversely related if y is
  a multiple of 1/ x
• 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 xy k
• We say that x and y vary jointly if y is a
  multiple of two or more variables.
• 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
9-2 Examples

• Direct variation Given y varies directly as x,
  and when x 1.5, y 3 , find y when x 4.
• 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 kx 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.
10
9-2 Examples

• Inverse Relationships If y varies inversely as
  x, and when x 5, y 8, what is y when x -2 ?
• Remember that in inverse relationships, y k /
  x.
• Use the given information 8 k / 5.
• Cross multiplying gives 8 5 k 1, or 40
  k.
• State the equation y 40 / x.
• Solve using the new equation. y 40 / -2. This
  gives y -20, the solution.
• Alternatively, you could use the initial equation
  xy k.
• This still gives 8 5 k, k 40. It would also
  give -2y 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.
11
9-2 Examples

• 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?
• Remember that in joint variation, y kxz.
• Use the given information 12 k(-2)(4).
• Solving for k gives 12 k(-8) , 12/-8 k , k
  -1.5
• State the equation y (-1.5)xz .
• Solve using the new equation. y (-1.5)(5)(2).
  This gives y -15, the solution.

12
9-2 Problems

• Use x 4, y -12, and z 5 to solve the
  following
• 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
• For the following, state if the relationship is
  direct, inverse, or join variation, and find the
  constant of variation, k
• XY 4 C. X Z -2 Y
• 12Y X 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.
13
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
  (33)/(35) 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)(x3) xx(x-2)
(x3) x
x(x-2)(x3) xx(x-2)



14
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
3x x2
(x1) (x2)
3x(x1) X2(x2)
As long as B ? 0 and D ? 0. For example




4x x3
(x7) (x-2)
A B
C D
4x(x-2) X3(x7)


As long as B ? 0, C ? 0 D ? 0. For example


15
9-3 Example

• Give the simplified form of each rational
  equation

x2 x5
3x x
3xx2 x(x5)
3x3 x(x5)
3x3 2 x(x5)
3x2 x5





x x-2
x2-4 x
x(x2-4) x(x-2)
(x2)(x-2) (x-2)
(x2)(x-2) (x-2)
x(x2-4) x(x-2)
x2-4 (x-2)






x2

x15 12x4
Note 2/24 1/12 x/x5 1/x4
x15 6x
4x4 2x
2x(x15) 6x4x4
2x(x15) 24x5
2x(x15) 24x5 4





(x4)(x-3) -3x4
x-3 2x8
2x8(x4)(x-3) -3x4(x-3)
2x8(x4)(x-3) -3x4(x-3)
2x4(x4) -3




16
9-3 Problems
7x x5
3x(x2) x-10

• Multiply
• Multiply and simplify
• Divide
• Divide and simplify

x2 (y2) x7
y5 4y2(y2)

xy3 y-8
3x2 y4

y(x4) y4(y-5)
x2(x4) x4

21x2(x2) (x-10)(x5)
y3(y2) 4x5
xy3(y4) 3x2 (y-8)
x2 y3(y-5)
2.
3.
1.
4.
17
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 3
3 8
18 38
33 38
8 24
9 24
17 24






x2 4x2
3 7x
7(x2) 74x2
4x3 4x7x
7x14 28x2
12x 28x2
19x14 28x2






22a2c 21ab3c 77ab2c
2a 7b2
3abc 11ac
2a11ac 7b211ac
3abc7b2 11ac7b2
22a2c 77ab2c
21ab3c 77ab2c






18
9-4 Examples

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

3xy 2(y1)
y 15x2
3xy15x2 2(y1)15x2
y2(y1) 15x22(y1)
45x3y 30x2(y1)
2y(y1) 30x2(y1)
Since these denominators have nothing in common,
we multiply the numerator and denominator of each
fraction by the opposite denominator.
Now we have a common denominator, we can
simplify, then subtract.
45x3y 2y(y1) 30x2(y1)
Sometimes only one portion of the rational
equations needs adjusting.
3x 2y
x2 4y3
3x2y2 2y2y2
x2 4y3
6xy2 4y3
x2 4y3
6xy2 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.
19
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
1x2 2yx2
y2y x2 2y
1x2 2yx2
y2y x2 2y
2x22y2 2x2y



2x22y2 2x2y
6y 30y - 4
x
4 6y
56y 16y
4 6y
56y 16y
4 6y
30y - 4 6y
5
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.
20
9-4 Problems

• Find the Least Common Denominator
• Solve the following equations
• Solve the following, using sections 9-3 and 9-4
  material.

y 4x2y
y 3xy3
2x 3x(y2)
y2 3xy6x
x y 10x2
4x2 5xy
12x2y3 , 3xy6x , 10x2y
x2 19x2
y 4(x3)
x4 72xy
y x
21
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.

22
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 lt 5 x gt - 5/3
23
9-5 Examples

• Solve the following rational equation for x.

3 2(x1)
1 15x
3 2(x1)
1 15x
3 2
1 15
30x(x1)
30x(x1)


30x
30(x1)


Here we see some common terms between the
numerator and denominator that can be canceled
out.
The LCD is 30x(x1). Multiply the LCD by both
side of the equation.
Next, multiply the 30x and 30(x1) into the
rational expressions.
45x
2x2
90x 2
30(x1) 15
45x
2(x1)
45x-2x-2
0
Plugging the solution x2/43 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.
Simplify the fractions, and distribute the
expression in the numerator in the right equation.
Now solve using algebra.
43x-2
0
43x
2
2 43
x
24
9-5 Examples

• 2. Solve the following rational inequality for x

x 7(x2)
5 x1
x 7(x2)
x x1
x(x1) 7(x2)(x1)
x7(x2) 7(x2)(x1)
lt
lt
0
lt
0
Start by subtracting the right side to the left.
First, get a common denominator. The LCD is
7(x2)(x1)
Next, simplify.
x(x1) x(x2) 7(x2)(x1)
x2x x2 2x 7(x2)(x1)
-x 7(x2)(x1)
x (x2)(x1)
lt
0
lt
0
lt
0
gt
0-7
Combine the two parts now that there is a common
denominator.
Multiply each side by -7 to remove these
constants. Dont forget to change inequality.
Plugging x-3 into the inequality gives
-1.5 gt 0 which is false. X -1.5 gives 6 gt 0
which is true X -0.5 gives -0.66 gt 0 which is
false. X 1 gives .166 gt 0 which is true.
x (x2)(x1)
gt
0
Thus, the final solution is -2 lt x lt -1 and x gt 0.
From here, we know that x is zero or undefined at
x0, -2, and -1. Using this information, use a
number line to test ranges.
?
?
X
X
-2 -1 0
25
9-5 Problems

• Solve the following rational equations
• Solve the following rational inequalities

z -1 z2
z-1 z3
x x2
1 2
3 5x5
1 x1
2 5




x 2
z 1
x -2
1 x1
x 2
x-2 x
4 x4
gt
lt
x lt -2 and -1ltxlt1
-4 ltxlt -2 and 0ltxlt4