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! :)

Guest May 22, 2019

#1**+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

CPhill May 22, 2019