SOLUTION: Find the vertical asymptote(s) and/or holes for R(x)=4/(x² -25)(x+2). I have the vertical asymptotes as x=5,x=-5 and x=2. Is this correct? As for holes I am lost, haha,

Algebra ->  Functions -> SOLUTION: Find the vertical asymptote(s) and/or holes for R(x)=4/(x² -25)(x+2). I have the vertical asymptotes as x=5,x=-5 and x=2. Is this correct? As for holes I am lost, haha,       Log On


   

Question 953557: Find the vertical asymptote(s) and/or holes for R(x)=4/(x² -25)(x+2).
I have the vertical asymptotes as x=5,x=-5 and x=2. Is this correct? As for holes I am lost, haha,
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
What, this? R(x)=4/((x² -25)(x+2))

Trying to render, R%28x%29=4%2F%28%28x%B2+-25%29%28x%2B2%29%29

Do you mean this?
R%28x%29=4%2F%28%28x%5E2-25%29%28x%2B2%29%29

Vertical asymptotes are x=5, x=-5, x=-2.
Factors of the denominator do not occur in the numerator, so no holes.

Finding your way: y=%28x-r%29%2F%28%28x-c%29%28x-r%29%29 is equivalent to y=%28%28x-r%29%2F%28x-r%29%29%281%2F%28x-c%29%29.