How do you find the vertical, horizontal or slant asymptotes for #(x^2 + 3x - 18) / (x^2- 36)#?

1 Answer
Feb 27, 2016

Answer:

Oblique: N/A
Horizontal: #1#
Vertical: #6#

Explanation:

The very first thing I do when I see a rational function, like #(x^2+3x-18)/(x^2-36)# is to factor. Factoring breaks down a problem and allows me to see the components of it and maybe even lets me simplify the problem. For this one, after factoring it, we get #((x-3)(cancel(x+6)))/((x-6)(cancel(x+6)))#. Now, the #x+6#s cancel out (that means that #0=x+6# or #x=-6# is a hole, BTW), which leaves us with #(x-3)/(x-6)#.

There are a few things that we can learn from this. One is that the vertical asymptote is #0=x-6#, or #x=6#.We know this because an asymptote is a value that the function can get very very close to but can never equal. The reason why is that that value will make the equation divide by zero, which can never happen. Because #6# is the value that makes #x=0#, #x# can never equal that.

Another thing that we can learn is the horizontal asymptote. Now, when the "power" or exponents are equal, like right now, then the horizontal asymptote is equal to the ratio of the coefficients, which in this case is #1/1#, or just #1#. If there is a higher power on bottom than on top then the horizontal asymptote is #y=0#. In the cases where the numerator is higher than the denominator, then there is no asymptote.

The only time that there is an oblique (slant) asymptote is when the numerator is a higher power than the denominator, which does not occur in this case.

So, the vertical asymptote is #x=6#, the horizontal asymptote is #y=1#, and there is no oblique asymptote, and there's a hole at #x=-6#. Let's graph the equation and see if we're right.
graph{(x^2+3x-18)/(x^2-36)}
And we were! Nice job!