# How do you identify all asymptotes or holes for y=(x^3-1)/(x^2+2x)?

Jan 19, 2018

There is no hole. Vertical asymptotes are $x - 0$ and $x = - 2$ and onlique or slanting asymptote is $y = x$

#### Explanation:

A funcion of the type $\frac{f \left(x\right)}{g \left(x\right)}$ is not defined for values for which $g \left(x\right) = 0$.

But let $g \left(x\right) = a \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$ and if any of $\left(x - \alpha\right) , \left(x - \beta\right)$ or $\left(x - \gamma\right)$ - say $\left(x - \beta\right)$ is a factor of $f \left(x\right)$ too, we can still get the value of function as $\left(x - \beta\right)$ cancels out.

We call a hole at $x = \beta$ , as on both sides of $x = \beta$, $\frac{f \left(x\right)}{g \left(x\right)}$ is not only defined, the value of the function $\frac{f \left(x\right)}{g \left(x\right)}$ can also be find using limits.

Observe that $y = \frac{{x}^{3} - 1}{{x}^{2} + 2 x}$ can be written as ((x-1)(x^2+x+1))/(x(x+2) and their is no common factor between numerator and denominator. As such their is no hole.

Further as we have $x \left(x + 2\right)$ in denominator when $x \to 0$ or $x \to - 2$, from left or right function, the function $y \to \pm \infty$, hence, we have vertical asymptote at $x = 0$ and $x = - 2$.

We have horizontal asymptote if degrees of $f \left(x\right)$ and $g \left(x\right)$ are equal and an obliique or slanting asymptote if degree of $f \left(x\right)$ is just one more than that of $g \left(x\right)$.

Hence here we have an obliique or slanting asymptote. As $y = \frac{{x}^{3} - 1}{{x}^{2} + 2 x} = \frac{x - \frac{1}{x} ^ 2}{1 + \frac{2}{x}}$ and as $x \to \infty$, we have $y = x$ and we have obliique or slanting asymptote as $y = x$.

graph{(x^3-1)/(x^2+2x) [-41.33, 38.67, -22.32, 17.68]}