# Rational Functions

• Apr 5th 2010, 01:01 PM
Hellooo
Rational Functions
Hi,
If you were given, say: x-intercept of -4, y-intercept of -2, vertical asymptote x=2, and horizontal asymptote y=1. How do you find the equation for a rational function with these features?
• Apr 5th 2010, 02:55 PM
skeeter
Quote:

Originally Posted by Hellooo
Hi,
If you were given, say: x-intercept of -4, (x+4) in the numerator

y-intercept of -2, y = -2 when x = 0

vertical asymptote x=2, (x-2) in the denominator

and horizontal asymptote y=1. degree of numerator = degree of denominator and ratio of the leading coefficients = 1

$y = \frac{x+4}{x-2}$
• Apr 6th 2010, 03:25 PM
Hellooo
Thanks, but can you show the steps in how you reached the answer?
• Apr 9th 2010, 05:36 PM
mr fantastic
Quote:

Originally Posted by Hellooo
Hi,
If you were given, say: x-intercept of -4, y-intercept of -2, vertical asymptote x=2, and horizontal asymptote y=1. How do you find the equation for a rational function with these features?

The simplest model to use is $y = \frac{a}{x - b} + c$. Your job is to find the values of a, b and c using the given information. You will get an answer equivalent to skeeter's.

In fact, the only information required is vertical asymptote x=2, horizontal asymptote y=1 and y-intercept of -2. The information x-intercept of -4 can be used to check your answer.
• Apr 9th 2010, 07:33 PM
AllanCuz
Quote:

Originally Posted by Hellooo
Thanks, but can you show the steps in how you reached the answer?

To supplement to Mr. Fantastic's information: horizontal asymptotes occur in the numerator and vertical asymptotes occur in denominator. If you recall a function cannot cross an asymptote (always in the case of vertical, most times true in the case of horizontal). Which means that the function cannot be a certain value there.

Thus, if a vertical asymptote of 2 exists we must have (x - 2) in the denominator.

Also recall that at the x-intercept y = 0, thus we must have (x+4) in the numerator because our x-intercept of -4 makes this equation 0.