Question

How do you find the asymptotes for `f(x)=x^2/(x+5)`?

Answer 1

The vertical asymptote is `x=-5`
The slant asymptote is `y=x-5`
There is no horizontal asymptote

Explanation:

The vertical asymptotes are calculated by performing the limits

`lim_(x->-5^(-))f(x)=lim_(x->-5^(-))x^2/(x+5)= 25/(0^-) = -oo`

`lim_(x->-5^(+))f(x)=lim_(x->-5^(+))x^2/(x+5)= 25/(0^+) = +oo`

The vertical asymptote is `x=-5`

We perform a long division to calculate the slant asymptote

`color(white)(aaaa)` `x+5` `color(white)(aaaa)|` `x^2+0x+0` `|` `color(white)(aaaa)` `x-5`

`color(white)(aaaaaaaaaaaaa)` `x^2+5x`

`color(white)(aaaaaaaaaaaaaa)` `0-5x`

`color(white)(aaaaaaaaaaaaaaaa)` `-5x-25`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `0+25`

Therefore,

`f(x)=x-5+25/(x+5)`

`lim_(x->-oo)f(x)-(x-5)=lim_(x->-oo)25/(x+5)=0^-`

`lim_(x->+oo)f(x)-(x-5)=lim_(x->+oo)25/(x+5)=0^+`

The slant asymptote is `y=x-5`

To determine the horizontal asymptote, we calculate

`lim_(x->-oo)f(x)=lim_(x->-oo)x^2/(x+5)=-oo`

`lim_(x->+oo)f(x)=lim_(x->+oo)x^2/(x+5)=+oo`

There is no horizontal asymptote

graph{(y-x^2/(x+5))(y-x+5)=0 [-58.5, 58.53, -29.27, 29.28]}