# How do you graph f(x)=x^2/(x+1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 7, 2018

graph{x^2 / (x+1) [-10, 10, -5, 5]}

#### Explanation:

You must know that this function means that $y = {x}^{2} / \left(x + 1\right)$

To find where $x = 0$, just plug it into the above equation.
$y = \frac{0}{0 + 1} \implies y = 0$

To find where $y = 0$, just plug it into the above equation.
$0 = {x}^{2} / \left(x + 1\right) \implies x = 0$

To find where we have vertical asymptotes you must find for some "explosive" point. In this case, we have the division by zero (we could have $\ln \left(0\right)$ too, for example).

The point where the function blows up is in $x = - 1$. So, here we have a vertical asymptote. Imagine if you plug value near to -1 "to the right". You would have a positive value divided by a very small positive value, it means that the function "goes" to infinity if you come from the right side. If you take very near values by the left you'll have a positive divided by a negative value, so it goes to negative infinity.

Horizontal asymptotes. You must see the limit of the function.
${\lim}_{x \to \setminus \infty} {x}^{2} / \left(x + 1\right) = {\lim}_{x \to \setminus \infty} \frac{2 x}{1} = \setminus \infty$
${\lim}_{x \to - \setminus \infty} {x}^{2} / \left(x + 1\right) = {\lim}_{x \to - \setminus \infty} \frac{2 x}{1} = - \setminus \infty$

So we don't have horizontal asymptotes.