M17 t1 notes

Definition:A rational function is a function that can be

written in the form of a polynomial divided by a polynomial,

.0)( , )(

)(xqwhere

xq

xp

Examples:

1)(

x

xxf

3

3)(

xxg 52

1)(

xx

xxh

Because rational functions are expressed in the form of a fraction, the denominator ofa rational function cannot be zero, since division by zero is not defined.

Domain:The domain of a rational function is the set of

all real numbers, except the x-values that make the denominator zero.

Example:

The domain is the set of all real numbers except 0, 2, & -5.

)5)(2(

97)(

2

xxx

xxxf

Graphing Rational Functions

372

12

xx

xy

Step 1: Identify zeros (x-intercepts) of the function.The zeros of this function are found by setting y to 0 and solving for x:

We see that when we do this, we find that the numerator determines our zeros. In this case, our zero is found by solving x −1=0, so x = 1 is our zero.

372

10

2

xx

x

Step 2: Identify our vertical asymptotes.We know that this function is undefined when the denominator is 0. So, let us find for which x-values the denominator is equal to zero. Our graph will contain vertical asymptotes at these x-values. We want to solve the equation 2x2 +7x+3=0.

0372 2 xx

0)3)(12( xx

0)12( x 0)3( x

2

1x 3x

So, our graph contains two vertical asymptotes, which represent values of x that our function does not contain in its domain. Our graph will tend towards these x-values, but will never pass through them.

Step 3: Identify our horizontal asymptote.

By definition, a horizontal asymptote is a y-value that a graph tends toward as the x values approach their extremes (∞ and - ∞). (Note that this doesn’t mean a graph can’t cross its horizontal asymptote for relatively small x-values.)

We identify our horizontal asymptote by considering what y-value our graph approaches as x → ∞ and x → −∞ . We will do this by dividing every term in the function by the highest degree term in the denominator:

2

2

222

2

22

2

2

2

23

27

1

21

21

23

27

22

21

2

21

21

372

1

xx

xx

xxx

xx

xxx

x

xxx

x

Notice that when x → ∞ , each term that contains an x in the denominator after our last step approaches 0, since we’re dividing constants by more and more. So, as x → ∞ , . Similarly, as x → −∞ , y → 0 . So, our horizontal asymptote is y = 0 .0

001

00

y

Notice that the degree of the numerator is less than that of the denominator. If this is the case, the horizontal asymptote is y = 0 .

Step 4: Check whether graph crosses horizontal asymptote.

Since we can cross horizontal asymptotes for relatively small values of x, we will check

whether we do so in this example. To cross our horizontal asymptote would mean that y

= 0 for some value(s) of x. So, we want to solve the equation .

We found in Step 1 that x = 1 is the solution to this equation, so our graph does cross our

horizontal asymptote when x = 1.

372

10

2

xx

x

3

1

3)0(7)0(2

102

y

3

1

Step 5: Find y-intercept.

We find our y-intercept by setting x to be 0 and solving for y. So, since

our y-intercept is .

Step 6: Graph lies above or below x-axis.

Our graph is split up into three sections, since we have two vertical asymptotes. One

section is all x-values less than -3, the next section is x-values between -3 and ,

and the third section is all x-values greater than . The third section has two

sections in itself, since we have a zero in that interval. We want to determine where

our graph lies in relation to the x-axis in each of these four sections. We will pick a

value of x in each of these sections to determine the sign of the function in each of

these sections:

2

1

2

1

Section Test Point Value of function Above or below the x-axis

x < -3 x = -4 y < 0 Below

-3 < x < x = -1 y = 1 Above

< x < 1 x = 0 y = < 0 Below

x > 1 x = 2 y = 0.04 Above

2

1

2

1

3

1

714.0

Step 7: Graphing our function.

We put all of the elements we just found in the previous 6 steps together and obtain the following graph:

3x 2

1x

Let’s look at another one…

Graph the rational function:

1)Identify the horizontal asymptote Since the numerator and the denominator are both

the same degree (2), you find the horizontal asymptote by dividing the leading coefficients.

y= 1/1 = 1

6

42

2

xx

xy

Step 2Let y = 0 to find the possible x intercepts

x

x

x

xx

x

2

4

40

6

40

2

2

2

2

Step 3To find the vertical asymptotes, set the denominator

equal to zero and solve

Notice that the -2 showed up as a asymptote AND an intercept. That means that something is going on

around -2.Look at this….

2.................3

02......03

)2)(3(0

60 2

xorx

xorx

xx

xx

Step 3bIf you factor the numerator and the

denominator, you are able to reduce…

)3)(2(

)2)(2(6

42

2

xx

xxy

xx

xy

When you can reduce, this creates a hole. Think of asymptotes like walls and holes like puddles…you can’t walk through a walk, but you can jump over a puddle.

When you see duplicate values, that means there is a hole and not really an asymptote.

You can even look for holes first!

Step 4Find the y intercept by letting x = 0

3

2

6

4

600

402

2

y

Start putting it togetherWhen we put the asymptotes, intercepts, and hole on the graph it’s easy to see where the graph goes…

Draw in the graph…**Rational f(x)’s “flip-flop” over the

asymptotes**

Using the TIYou can use the calculator to graph these, just make

sure you enter the function correctly! The numerator and

the denominator need there own parenthesis!

You will need to enter: (x^2 -4) / (x^2 – x -6)

You will be able to see the vertical asymptotes ONLY! They will appear as a line on the graph.

The rest is up to you

Practice Problem:Graph the function . Find all asymptotes and x- & y-intercepts.

12

22

xx

xy

x-intercept: x = 2y-intercept: y =Vertical asymptotes: x = -3 & x = 4Horizontal asymptote: y = 0

61