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

1 Answer
Jun 18, 2018

Answer:

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.