How do you find the asymptote(s) or hole(s) of #f(x) = (x+3)/(x^2-9)#?

2 Answers
Jun 6, 2018

Answer:

asymptote at #x=3#

hole at #x=-3#

Explanation:

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

Factor the denominator, it is the difference of squares:

#f(x) = (x+3)/((x+3)(x-3))#

The factor(s) in the denominator that will cancel out, in this case #x+3# are holes.

The factor(s) in the denominator that will not cancel out, in this case #x-3# are asymptotes.

Here is the graph, it shows the asymptote, but not the hole at #x=-3#

graph{(x+3)/(x^2-9) [-6.67, 13.33, -4.44, 5.56]}

Jun 6, 2018

Answer:

VA: #x=3#

HA: #y=0#

Explanation:

We have the following expression

#(x+3)/color(blue)((x^2-9))#

What I have in blue is a Difference of Squares of the form

#a^2-b^2#, where #a=x# and #b=3#, which is factored as

#(a+b)(a-b)#

Now, our new expression is

#(x+3)/color(blue)((x+3)(x-3))#

which simplifies to

#1/(x-3)#

Vertical asymptotes are found by setting the denominator equal to zero, because this is the value for which the function is undefined.

#x-3=0#

#=>color(red)(x=3)# (Vertical Asymptote)

Horizontal asymptotes are found by comparing the degrees. Since there is a higher degree in the denominator, we have a horizontal asymptote at

#=>color(darkviolet)(y=0)# (Horizontal Asymptote)

We have a vertical asymptote at #x=3#, and a horizontal asymptote at #y=0#.

Hope this helps!