Graphing Rational Functions

19 - 2 Graphing Rational functions

Try the quiz at the bottom of the page!

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Steps for Graphing Rational Functions y = f(x)/g(x)

1) Find the zeros of g(x). These will be the vertical asymptotes unless it's also a zero of f(x)!

2) Find the horizontal asymptotes by studying

and

. If

, then y=c is a horizontal asymptote.

3) Find the zeros of f(x). These will be the x-intercepts unless it's also a xero of g(x)!

4) If you have a common zero of f(x) and g(x), this represents a hole in the graph! Reduce the fraction and plug in the x value. This is where the hole exists!

5) Replace x with zero. If the denominator is not zero, you have found the y-intercept!

6) Do a sign analysis in each interval seperated by vertical asymptotes and x-intercepts. This tells you where each section of the graph appears!

7) Use the vertical and horizontal asymptotes to help you sketch the graph. If x = c is a vertical asymptote, then the graph approaches infinity as it nears the asymptote in a region where the function is positive, and it approaches negative infinity as it nears the asymptote in a region where the function is negative. (Note: vertical asymptotes can not be crossed because they describe where the graph is undefined. Horizontal asymptotes may be crossed as they describe only what happens to the graph as x gets very large or very small!)

Sample Problem.

1) Graph the following function:

a) First find the vertical asymptotes by finding the zeros of the denominator! That would be x = -2

b) Now find the horizontal asymptotes by finding the limits on infinity for the function.

and Remember to find the limit divide by the highest power in the denominator!

The graph should look like this:

red lines are the asymptotes!

c) Find the zeros of the numerator! It's x = -2/3. This is not a zero of the denominator, so it is the x-intercept. The only x-intercept!

d) Find the y-intercept (if any), by replacing x with zero. You get 2/4 = 1/2

e) Do a sign analysis for the areas seperated by any vertical asymptotes and zeros.
The graph now looks like:

The graph must appear in the top left hand corner because in that area, the function is always positive. It must approach the vertical and horizontal asymptote. The area between the vertical asymptote and the x-intercept is negative. It must go down into the negative area again approaching the vertical asymptote. The area to the right of the x-intercept is positive. Therefore, it's above the x-axis and approaches the horizontal asymptote. To find out if it crosses the asymptote, set the function = to 3/2 and solve for x. You will discover that the x's drop out and you end up with 12 = 4. No solution, therefore, it doesn't cross the horizontal asymptote!
The final graph looks like:

Sample problem #2

2) Graph the following function:

a) Find the zeros of the denominator after you simplified. The vertical asymptote is x = -3.

b) Find the limits on infinity of the function after you simplify.

The horizontal asymptote is y = 1.

The graph looks like this:

c) Find the x-intercept by finding the zeros of the numerator. x = 0

d) Find the y-intercept by replacing the x value with 0. y-int is 0.

e) Find the holes in the graph. The hole appears at the factor that was canceled. In this case, at x = 3. To find the y value, plug x=3 into the simplified function and get 3/6 = 1/2. The hole is at (3, 1/2)

The graph now looks like this:

Note the sign analysis for each area of the graph. In the upper left corner, the y values are + which means the graph is above the x-axis approaching the vertical and horizontal asymptote. The area to the left of the zero is negative and approaches the vertical axis downward. The area to the right of the zero reverts back to positve and approaches the horizontal asymptote. It will not cross the asymptote because setting 1 = x/(x+3) yields no solutions! Here's what it looks like:

Note: the hole is marked by the purple cross on the graph!

Sample problem #3

3) Graph the following function:

a) Factor the function into f(x) = (x - 5)/(x - 2)(x + 2). Find the vertical asymptotes by finding the zeros of the denominator. None of the factors cancel, so there are two vertical asymptotes at : x = 2 and x = -2.

b) Find the horizontal asymptote by finding the limits on infinity of the function.

Therefore, the horizontal asymptote is y = 0

c) Find the x-intercept by finding the zeros of the numerator. x = 5 is the x-intercept.

d) Find the y-intercept by replacing x with 0. (0 - 5)/(0 - 4) = 5/4
The graph looks like this:

e) Find the holes in the graph. No holes in the graph because there were no factors that canceled. Do the sign analysis and the graph looks this way:

The sign analysis goes between each vertical asymptote and the x-intercepts.
The final graph looks like this:

This section really "rules", but alas it comes to an end! On to the sample test or back to the previous section?

Current quizaroo # 19b

a) -2; b) -3; c) both -2 and -3; d) 0; e) doesn't have one
: a) 6; b) -12; c) -1/2; d) both 6 and -12; e) doesn't have one
3) Name the vertical asymptotes, if any.: a) x = -3; b) x = -2; c) x = 4; d) x = 4 and x = -3; e) x = -2 and x = 4

4) Name the horizontal asymptotes, if any.: a) y = 0; b) doesn't have one; c) y = 3; d) y = 4; e) y = 1

5) Name any holes in the graph, if any.

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