SOLUTION: f(x)=(2x-4)/(x^2-6x-16): What is the vertical asymptote, horizontal asymptote, x-intercept, and y-intercept? Could anyone explain how to do this kind of problem to me as well? I

Algebra ->  Rational-functions -> SOLUTION: f(x)=(2x-4)/(x^2-6x-16): What is the vertical asymptote, horizontal asymptote, x-intercept, and y-intercept? Could anyone explain how to do this kind of problem to me as well? I      Log On


   

Question 426804: f(x)=(2x-4)/(x^2-6x-16): What is the vertical asymptote, horizontal asymptote, x-intercept, and y-intercept?
Could anyone explain how to do this kind of problem to me as well? I have a test tomorrow!
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
We can factor the function as:
f%28x%29=%282x-4%29%2F%28x%5E2-6x-16%29+=+2%28x-2%29%2F%28x%2B2%29%28x-8%29
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%2Fx%5E2+-%3E+2%2Fx
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%28300%2C200%2C-20%2C20%2C-8%2C8%2C%282x-4%29%2F%28x%5E2-6x-16%29%29