Answer
vertical asymptote $x=3$
no horizontal asymptote.
slant asymptote $y=x+5$
See graph.
Work Step by Step
Step 1. Rewrite the function as
$y=\frac{x^2+2x-3}{x-3}=\frac{x^2-3x+5x-15+12}{x-3}=x+5+\frac{12}{x-3}$
Step 2. The vertical asymptote can be found as $x=3$
Step 3. There is no horizontal asymptote.
Step 4. A slant asymptote can be found as $y=x+5$
Step 5. See graph.
