# How do you find the vertical, horizontal or slant asymptotes for #y= (x-4)/(x^2-8x+16)#?

##### 1 Answer

May 23, 2016

vertical asymptote x = 4

horizontal asymptote y = 0

#### Explanation:

The first step here is to factorise and simplify the function.

#rArr(cancel((x-4)))/(cancel((x-4))(x-4))=1/(x-4)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 4 = 0 → x = 4 is the asymptote.

Horizontal asymptotes occur as

#lim_(xto+-oo) , y to 0 # divide terms on numerator/denominator by x

#(1/x)/(x/x-4/x)=(1/x)/(1-4/x)# as

#xto+-oo ,yto0/(1-0)#

#rArry=0" is the asymptote"# slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0,denominator-degree 1).Hence there are no slant asymptotes.

graph{1/(x-4) [-10, 10, -5, 5]}