How do you graph #y=3/(x^3-27)# using asymptotes, intercepts, end behavior?

1 Answer
Jan 13, 2018

Answer:

Read below.

Explanation:

We first try to find the intercepts.

To find the #x# intercepts, we set #y=0#.
#0=3/(x^3-27)#
#0=3# Clearly there is no solution. Therefore, we know that there is no #x# intercept.

To find the #y# intercepts, we set #x=0#.
#y=3/(0^3-27)#
#y=3/(-27)#
#y=-1/9# The point #(0,-1/9)# is the only x intercept.

To figure out its end behavior, we have to figure out the following:

What is #lim_(x->oo)3/(x^3-27)#?
What is #lim_(x->-oo)3/(x^3-27)#?

What is #lim_(x->oo)3/(x^3-27)#?
Let's use logic here.
As #x# gets really, really, large, the denominator will get really, really, large. Subtracting it with 27 is too insignificant to affect the result.

Now, as the denominator get larger and larger, the fraction gets closer and closer to 0.
#lim_(x->oo)3/(x^3-27)=0#

What is #lim_(x->-oo)3/(x^3-27)#?

As #x# gets really, really, small, the denominator will be a very, very, small negative number. Subtracting it with 27 is too insignificant to affect the result.

Now, as the denominator gets smaller and smaller, the fraction gets closer and closer to 0. (Remember that the denominator never really reaches 0.)
#lim_(x->oo)3/(x^3-27)=0#