Question

What are the asymptote(s) and hole(s), if any, of `f(x)= (x+3)/(x^2-9)`?

Answer 1

Hole at `color(red)((-3, -1/6)`

Vertical asymptote: `x= 3`

Horizontal asymptote: `y= 0`

Explanation:

Given `f(x)= (x+3)/(x^2-9)`

Step 1: Factor the denominator, because it's a difference of square

`f(x) = (x+3)/((x+3)(x-3)) hArr f(x)=cancel(x+3)/(cancel (x+3)(x-3)) " " " " " " " " " " " " " " " " " " " " " " " " " " " hArrcolor(blue)( f(x)= 1/(x-3))`
Because the function reduce to equivalent form , we have a hole on the graph at

`x+3 = 0 hArr x= -3`

`y_(value)= f(-3)= 1/(-3-3) hArr f(-3) = -1/6`
Hole at `color(red)((-3, -1/6)`

Vertical asymptote: Set denominator equal to zero

`x-3= 0 hArr x= 3`

Vertical asymptote : `x= 3`

Horizontal asymptote:
`f(x) = (1x^0)/(x-3)`

Because the degree of the numerator is LESS than the degree of the denominator, the horizontal asymptote is

`y= 0`