+0

Writing the equation for this rational function graphed below?

0
43
3

So I know that the verticle asymptote is at x = -2, the horizontal asymptote at y = 1, and the point discontinuity is at (0,3).

So far, I have $${x - 3 \over (x + 2)(x - 3)}$$. I'm not sure about what to do next?

Thank you! :)

May 22, 2019

#1
+101086
+3

If the horizontal asymptote = 1  then  the degree of the numerator = the degree of the denominator and the coefficients on the lead terms will be the same....

If we have a vertical asymptote at x = -2  then  ( x + 2)  must be a linear factor in the denominator

Note that   x = - 6 is a root....therefore  (x + 6)  must be a factor in the numerator

So.....the polynomial  that we are looking for is

x ( x + 6)

_______

x (x + 2)

When the x"s   are cancelled,  we end up with  (x + 6) /(x + 2) ....and when x = 0, y = 3

But x cannot = 0  because the original function is undefined at that point....so....the "hole" occurs at (0, 3)

Here's the graph  with the asymptotes and pertinent points : https://www.desmos.com/calculator/wge4wepw0i

May 22, 2019
edited by CPhill  May 22, 2019
#2
+1

Just wanted to clarify, will the horizontal asymptote always equal the degree of the numerator and denominator? :)

Guest May 22, 2019
#3
+101086
+2

Sorry.....I meant to say that if the horizontal asymptote  is y = 1.....the degree of the polynomials in the numerator and denominator is the same and their lead coefficients are also the same....

CPhill  May 22, 2019