Question

How do you graph `y=x^3/(x^2+7)` using asymptotes, intercepts, end behavior?

Answer 1

See explanation below

Explanation:

The domain of y is `D_y=RR`

The denominator is `(x^2+7)>0`, so we don't have a vertical asymptote.

As the degree of the numerator is `>` the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `x^3` `color(white)(aaaaaaaa)` `∣` `x^2+7`

`color(white)(aaaa)` `x^3+7x` `color(white)(aaaa)` `∣` `x`

`color(white)(aaaaa)` `0-7x`

So, `x^3/(x^2+7)=x-(7x)/(x^2+7)`

Therefore, `y=x` is an oblique asymptote

For the limits, we take the terms of highest degree in the numerator and the denominator

`lim_(x->+-oo)y=lim_(x->+-oo)x^3/x^2=lim_(x->+-oo)x=+-oo`

For the intercepts, when `x=0`, `=>` `y=0`

graph{(y-x^3/(x^2+7))(y-x)=0 [-7.9, 7.904, -3.956, 3.946]}