Question 426804
We can factor the function as:
{{{f(x)=(2x-4)/(x^2-6x-16) = 2(x-2)/(x+2)(x-8)}}}
We see immediately from this that there are 2 vertical asymptotes: x=-2, x=8, since f(x) -> {{{infinity}}} at these points
For large x, the function goes as {{{2x/x^2 -> 2/x}}}
This approaches, but never reaches 0, so the horiz. asymptote is: f(x)=0
x-intercept is when f(x)=0 -> 2(x-2) = 0 -> x=2
y-intercept: x=0 -> (0-2)/(-16), or x=1/8
This is the graph
{{{graph(300,200,-20,20,-8,8,(2x-4)/(x^2-6x-16))}}}