Question

How do you find vertical, horizontal and oblique asymptotes for `(x^3+5x^2)/(x^2-25)`?

Answer 1

The vertical asymptote is `x=5`
The oblique asymptote is `y=x+5`
No horizontal asymptote

Explanation:

The denominator is

`x^2-25=(x+5)(x-5)`

and the numerator is

`x^3+5x^2=x^2(x+5)`

Therefore,

`(x^3+5x^2)/(x^2-25)=(x^2cancel(x+5))/(cancel(x+5)(x-5))`

`=x^2/(x-5)`

Let `f(x)=x^2/(x-5)`

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

The vertical asymptote is `x=5`

The degree of the numerator is `>` than the degree of the denominator, there is an oblique asymptote.

Let 's do a long division

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

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

`color(white)(aaaaa)` `0+5x`

`color(white)(aaaaaaa)` `+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 oblique asymptote is `y=x+5`

graph{(y-(x^2)/(x-5))(y-x-5)(y+50x-250)=0 [-65.86, 65.94, -32.9, 32.9]}