Question

Hello, I need some assistance with this homework question, please? This is for my precalculus homework. Q10

Answer 1

.Explanation

We are given the function R(x)

`R(x)=(x^3-1)/(x^2+3x-4)`

So, to get the vertical, horizontal, and oblique asymptote

Let us begin with the vertical assymptote

Thus

`(x^(3)-1)/(x^(2)+3x-4)=((x^2+x+1)(x-1))/((x+4)(x-1))`
`((x^(2)+x+1)(x-1))/((x+4)(x-1))=((x^2+x+1))/((x+4))`

So, if we go by the definiation, the resulting function will be undefined when

`\begin{gathered} x+4=0 \\ so\text{ that} \\ x=-4 \end{gathered}`

Therefore, the vertical asymptote is x = -4

For the horzontal asymptote

`\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}`

In our case,

The degree of the numerator is greater than the denominator, therefore, There are no horizontal asymptotes

For the oblique asymptote.

Oblique asymptote is also known as the slant asymptote

To get the oblique asymptote, we will have to simplify the expression as

`(x^3-1)/(x^2+3x-4)=(13x-13)/(x^2+3x-4)+x-3`

The rational term approaches 0 as the variable approaches infinity

Thus, the oblique asymptote is y =x−3