Question

3
$f(x) = \frac{\cdots}{\cdots}$
hole:
no hole
vertical asymptote(s):
horizontal/slant asymptote:
(circle one)
domain:
$(-\infty, 3) \cup (3, \infty)$

Explanation:

Step1: Confirm simplified function

The given function factors to $f(x)=\frac{(x+3)(x-2)}{x-3}$, no common factors, so no holes.

Step2: Identify slant asymptote

Since degree of numerator ($2$) is 1 more than denominator ($1$), perform polynomial long division:
$$\frac{x^2+x-6}{x-3} = x+4 + \frac{6}{x-3}$$
As $x\to\pm\infty$, $\frac{6}{x-3}\to0$, so slant asymptote is $y=x+4$.

Step3: Verify domain

Denominator $x-3
eq0$, so $x
eq3$. Domain is $(-\infty,3)\cup(3,\infty)$.

Step4: Confirm vertical asymptote

Set denominator to 0: $x-3=0 \implies x=3$.

Answer:

• Hole: No hole
• Vertical Asymptote: $x=3$
• Slant Asymptote: $y=x+4$
• Domain: $(-\infty,3)\cup(3,\infty)$