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

Feb 15, 2018

Vertical asymptote: At $x = 1 , - 2$

Horizontal asymptote: At $y = 0$

X-intercept: None

Y-intercept: At $y = 1$

Holes: None

#### Explanation:

To find the holes and vertical asymptotes, we first study the denominator.

We have ${x}^{2} + x - 2$ in the denominator. Factorizing this will give us the vertical asymptotes.

${x}^{2} + x - 2$

${x}^{2} + 2 x - x - 2$

$x \left(x + 2\right) - 1 \left(x + 2\right)$

$\left(x + 2\right) \left(x - 1\right)$

The asymptotes are found when the answer is $0$.

So,

$\left(x + 2\right) \left(x - 1\right) = 0$

$x = 1 , - 2$

There are vertical aymptotes at $x = 1 , - 2$.

To find the horizontal asymptotes, we must look at the degree of the numerator ($n$) and the denominator ($m$).

If $n > m ,$ there is no horizontal asymptote

If $n = m$, we divide the leading coefficients,

If $n < m$, the asymptote is at $y = 0$.

Here, $n = 0$ and $m = 2$, so the horizontal asymptote is at $y = 0$.

Therefore, there is no x-intercept.

The y-intercept is found by taking all $x$ to be equal to $0$.

Doing so,

$- \frac{2}{{0}^{2} + 0 - 2}$

$- \frac{2}{-} 2$

$- \left(- 1\right)$

$1$

The y-intercept is at $y = 1$

There are no holes.