Question

How do you find the asymptotes for `f(x)=(4x)/(x^2-8x+16)`?

Answer 1

HA= `y= 0`
VA= x= +/- 4
No slant asymptote

Explanation:

factored form of `x^2-8x+16` is `(x-4) (x+4)` so therefor the VA are +/- 4
no slant asymptote because the degree of the numerator is not one more higher than the degree of the denominator.
Ha is 0 because the degree of x is higher on the bottom than the top.