Question

How do you find vertical, horizontal and oblique asymptotes for `y = (x-4)^2/(x^2-4)`?

Answer 1

vertical asymptotes x = ± 2
horizontal asymptote y = 1

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find equation/s let the denominator equal zero.

solve : `x^2 - 4 = 0 → (x-2)(x+2) = 0`

`rArr x = ± 2 " are the asymptotes "`

Horizontal asymptotes occur as `lim_(xto+-oo) f(x) to 0`

now numerator = `(x - 4)^2 = x^2 - 8x + 16`

and y = `(x^2-8x+16)/(x^2-4)`

divide all terms on numerator/denominator by `x^2`

`(x^2/x^2 -(8x)/x^2 + 16/x^2)/(x^2/x^2 - 4/x^2)=(1-8/x+16/x^2)/(1-4/x^2)`

As `xtooo , 8/x , 16/x^2 " and " 4/x^2 to 0`

`rArr y = 1/1 = 1 " is the asymptote "`

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(x-4)^2/(x^2-4) [-10, 10, -5, 5]}