# How do you graph y=1/(x^2-x-6)?

Aug 4, 2015

You find the intercepts and the asymptotes, and then you sketch the graph.

#### Explanation:

Step 1. Find the $y$-intercepts.

Set $x = 0$ and solve for $y$.

$f \left(0\right) = - \frac{1}{6}$

The $y$-intercept is at ($0 , - \frac{1}{6}$).

Step 2. Find the $x$-intercepts.

Set $y = 0$ and solve for $x$.

$0 = \frac{1}{{x}^{2} - x - 6}$

$0 = - 1$

This is an impossibility, so there is no $x$-intercept.

Step 3. Find the vertical asymptotes.

Set the denominator equal to zero and solve for $x$.

${x}^{2} - x - 6 = 0$

$\left(x - 3\right) \left(x + 2\right) = 0$

$x = 3$ and $x = - 2$

There are vertical asymptotes at $x = - 2$ and $x = 3$.

Step 4. Find the horizontal asymptote.

The degree of the denominator is less than the degree of the numerator, so

The horizontal asymptote is at $y = 0$ (the $x$-axis).

Step 5. Draw your axes and the asymptotes.

Step 6. Sketch the graph in each region defined by the asymptotes.

(a) The left hand region has the $x$-axis and $x = - 2$ as asymptotes.

$f \left(- 3\right) = \frac{1}{6}$.

The point at ($- 3 , \frac{1}{6}$) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) The right hand region has $x = 3$ and the $x$-axis as asymptotes.

$f \left(4\right) = \frac{1}{6}$.

The point at ($4 , \frac{1}{6}$) is in the first quadrant, so we have a "hyperbola" above the horizontal asymptote.

So we have a mirror-image hyperbola in the first quadrant.

(c) In the middle region, we have

$f \left(0\right) = - \frac{1}{6}$ and

$f \left(1\right) = - \frac{1}{6}$

$f \left(0.5\right) = - \frac{1}{6.25}$

The points at ($0 , - \frac{1}{6}$), ($0.5 , - \frac{1}{6.25}$), and ($1 , - \frac{1}{6}$) are all below the $x$-axis, so we have an "inverted parabola" between the vertical asymptotes.