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! :)
+128474 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
