How do you find the asymptotes for #y = (x-2)/( x^2-4)#?

2 Answers
Jan 28, 2016

Answer:

Vertical asymptote at #x=-2#, horizontal asymptote at #y=0#

Explanation:

First, we can simplify the expression.

#(x-2)/(x^2-4)=(x-2)/((x+2)(x-2))=1/(x+2)#

Here, the #x-2# has been cancelled out. That means that there is a removable discontinuity at #x=2#. A removable discontinuity is also known as a hole.

Our new function is #1/(x+2)# with a hole at #x=2#.

Vertical asymptotes:

The vertical asymptotes will occur when the denominator of the function equals #0#.

#x+2=0#

#x=-2#

There is a vertical asymptote at #x=-2#.

Horizontal asymptotes:

When the degree of the denominator is greater than the denominator of the numerator, the #x# axis or the line #y=0# is the function's horizontal asymptote.

We can check a graph:

graph{(x-2)/(x^2-4) [-10, 10, -5, 5]}

Jan 29, 2016

Answer:

The asymptote is #x=-2#

Explanation:

This is how I got that asymptote.
First, I factored everything as much as I can. It's like taking apart a puzzle to see every piece of it.
#x-2# is already as factored as it gets. However, #x^2-4# can be factored some more. Your first thought might be to factor it into this: #(x-2)^2#, but that would be wrong. #(x-2)^2# expanded is #(x-2)(x-2)#, which becomes #x^2-4x+4#. If we get an answer and then check it against the original and they are different, we did something wrong. #x^2-4x+4# does not equal #x^2-4#, so we have to figure something else out.

Now, #x^2-4# is actually a special case called "difference of squares", and from that I know that is can be factored to #(x+2)(x-4)#.

So now the equation looks like this: #(x-2)/((x+2) (x-2))# the #x-2#s will divide out, making a hole. That leaves #1/(x+2)#. Now, an asymptote is a line that the graph will get closer and closer to but never touch. This is because there is a value which will make the denominator equal to zero, and you cannot divide by zero, ever.

So let's find the value that makes #x+2=0#. Subtract #2# on both sides and we have #x=-2#. The line that the graph will never touch --the asymptote-- is #x=-2#.

We can always check our work by graphing
#(x-2)/((x+2) (x-2))# and seeing where that asymptote is.
graph{(x-2)/((x+2) (x-2))}