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

1 Answer
Nov 29, 2015

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