Question

Question #10591

Answer 1

VA: `x=-4` and `x=1`

Explanation:

Formatted question: Find VA (vertical asymptotes) of `f(x)=(4x+8)/(x^2+3x-4)`

First we need to factor the numerator and denominator of the given function:
`f(x)=(4(x+2))/((x+4)(x-1))`

Since there is nothing to cancel in the numerator and denominator (no holes), we know that the `x` values that make the denominator `0` will be vertical asymptotes.

So `x+4=0` and `x-1=0`

`therefore` `x=-4` and `x=1` are vertical asymptotes.