1. ## asymptote question

$y=\frac{(x^2-4)}{(x+2)(x-3)}$

Find all vertical, horizontal, and oblique asymptotes, and state if the function has any holes.

My first step was to simplify:

$y=\frac{(x-2)(x+2)}{(x+2)(x-3)}$

and cancel

$y=\frac{(x-2)}{(x-3)}$

Vertical asymptote at x=3. Horizontal asymptote at y=1. I don't think there is an oblique asymptote, because the degree of the numerator and denominator are equal. I believe there is a hole at $-2,\frac{4}{5}$

Is this correct? I guess my main question is, because I cancelled, does that eliminate the vertical asymptote x=-2 and replace it with the hole mentioned above?

2. Originally Posted by absvalue
$y=\frac{(x^2-4)}{(x+2)(x-3)}$

Find all vertical, horizontal, and oblique asymptotes, and state if the function has any holes.

My first step was to simplify:

$y=\frac{(x-2)(x+2)}{(x+2)(x-3)}$

and cancel

$y=\frac{(x-2)}{(x-3)}$

Vertical asymptote at x=3. Horizontal asymptote at y=1. I don't think there is an oblique asymptote, because the degree of the numerator and denominator are equal. I believe there is a hole at $-2,\frac{4}{5}$
Your work is OK. However, I'd format the coordinates of the hole as $\left( -2,\frac{4}{5}\right)$.
3. $y=\frac{(x-2)}{(x-3)}=\frac{1}{(x-3)}+1$
Therefore asymptotes are at $x=3$ and $y = 1$