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

1 Answer
Vikki
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#

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