Question

How do you find the vertical, horizontal and slant asymptotes of: `y =(x^3 - 64)/(x^2 + x - 20)`?

Answer 1

The vertical asymptote is `x=-5`
The slant asymptote is `y=x-1`
No horizontal asymptote

Explanation:

Let's factorise the denominator and the numerator

`x^2+x-20=(x+5)(x-4)`

`x^3-64=(x-4)(x^2+4x+16)`

Therefore, `y=(x^3-64)/(x^2+x-20)=(cancel(x-4)(x^2+4x+16))/((x+5)cancel(x-4))`

`y=(x^2+4x+16)/(x+5)`

As we cannot divide by `0`, so `x!=-5`

the vertical asymptote is `x=-5`

The degree of the numerator is `>` the degree of the denomitor

Let's do a long division

`color(white)(aaaa)` `x^2+4x+16` `color(white)(aaaa)` `∣` `x+5`

`color(white)(aaaa)` `x^2+5x` `color(white)(aaaaaaaa)` `∣` `x-1`

`color(white)(aaaaa)` `0-x`

`color(white)(aaaaaaa)` `-x+16`

`color(white)(aaaaaaa)` `-x-5`

`color(white)(aaaaaaaaa)` `0+21`

`:. y=(x^2+4x+16)/(x+5)=x-1+21/(x+5)`

So, `y=x-1` is a slant asymptote

`lim_(x->+-oo)y=lim_(x->+-oo)x^2/x=lim_(x->+-oo)x=+-oo`

So, no horizontal asymptote

graph{(y-((x^2+4x+16)/(x+5)))(y-x+1)=0 [-58.5, 58.56, -29.32, 29.23]}