Question

which statements are true about the graph of the function $f\left(x\
ight)=\frac{x^{2}-5x - 36}{x^{2}-81}$?\
\square there is a zero at $x=-4$.\
\square there is a zero at $x = 9$.\
\square there is a hole at $x=-9$.\
\square there is a hole at $x = 9$.\
\square there is an asymptote at $x=-9$.\
\square there is an asymptote at $x = 4$.

Explanation:

Step1: Factor numerator and denominator

Factor $x^2-5x-36$ and $x^2-81$:
$x^2-5x-36=(x-9)(x+4)$
$x^2-81=(x-9)(x+9)$
So $f(x)=\frac{(x-9)(x+4)}{(x-9)(x+9)}$

Step2: Identify holes

Holes occur where numerator/denominator share factors.
Cancel $(x-9)$; hole at $x=9$ (since $x=9$ makes the canceled factor 0).

Step3: Identify zeros

Zeros are roots of simplified numerator:
Set $x+4=0 \implies x=-4$ (check: $x=-4$ does not make denominator 0).

Step4: Identify vertical asymptotes

Asymptotes are roots of simplified denominator:
Set $x+9=0 \implies x=-9$.

Answer:

• There is a zero at $x = -4$.
• There is a hole at $x = 9$.
• There is an asymptote at $x = -9$.