Question

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

Answer 1

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]}