Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x)=(x+7)/(x^2-49)`?

Answer 1

vertical asymptote x = 7
horizontal asymptote y =0

Explanation:

The first step here is to factorise and simplify f(x)

`f(x) = (x+7)/((x+7)(x-7)) = cancel((x+7))/(cancel((x+7)) (x-7)) =1/(x-7)`

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

solve : x - 7 = 0 → x = 7 is the asymptote.

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

divide terms on numerator/denominator by x

`rArr (1/x)/(x/x - 7/x) = (1/x)/(1 - 7/x)`

as `x to+-oo , f(x) to 0/(1 - 0) = 0/1`

`rArr y = 0" is the asymptote "`

Slant asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here hence there are no slant asymptotes.
graph{(x+7)/(x^2-49) [-10, 10, -5, 5]}