Question

identify any vertical asymptotes and horizontal asymptotes. (enter your answers as comma - separated lists of equations. if an answer does not exist, enter dne.)
f(x)=\frac{-2(x + 3)(x - 8)}{(x - 5)(x - 8)}
vertical asymptotes

horizontal asymptotes

identify any holes.
((x,y)=(quad))

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to zero and solve for $x$, excluding common factors with the numerator. The denominator is $(x - 5)(x - 8)$. After canceling out the common factor $(x - 8)$ in the numerator and denominator, we consider the non - canceled factor. Setting $x-5=0$, we get $x = 5$.

Step2: Find horizontal asymptotes

Since the degrees of the numerator and denominator are the same (both degree 1 after canceling the common factor), the horizontal asymptote is given by the ratio of the leading coefficients. The leading coefficient of the numerator $-2(x + 3)$ is $-2$ and the leading coefficient of the denominator $(x - 5)$ is $1$. So $y=-2$.

Step3: Find holes

A hole occurs at the $x$ - value that makes a common factor in the numerator and denominator equal to zero. The common factor is $(x - 8)$. We find the $y$ - value of the hole by substituting $x = 8$ into the simplified function $f(x)=\frac{-2(x + 3)}{x - 5}$. $y=\frac{-2(8 + 3)}{8 - 5}=\frac{-2\times11}{3}=-\frac{22}{3}$.

Answer:

vertical asymptotes: $x = 5$
horizontal asymptotes: $y=-2$
$(x,y)=(8,-\frac{22}{3})$