# How do you graph (x^2-4)/(x^2-9)?

Jul 7, 2015

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

#### Explanation:

Step 1. Find the $y$-intercepts.

$y = f \left(x\right) = \frac{{x}^{2} - 4}{{x}^{2} - 9}$

$f \left(0\right) = \frac{{0}^{2} - 4}{{0}^{2} - 9} = \frac{- 4}{-} 9 = \frac{4}{9}$

The $y$-intercept is at ($0 , \frac{4}{9}$).

Step 2. Find the x-intercepts.

$0 = \frac{{x}^{2} - 4}{{x}^{2} - 9}$

${x}^{2} - 4 = 0$

${x}^{2} = 4$

x = ±2

The $x$-intercepts are at ($- 2 , 0$) and ($2 , 0$).

Step 3. Find the vertical asymptotes.

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

${x}^{2} - 9 = 0$

${x}^{2} = 9$

x = ±3

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

Step 4. Find the horizontal asymptote.

Both equations are of the second order, so we divide the coefficients of the ${x}^{2}$ terms.

$\frac{1}{1} = 1$

The horizontal asymptote is at $y = 1$.

Step 5. Draw your axes and the asymptotes.

The vertical asymptotes divide the graph into three regions of $x$s.

Step 6. Sketch the graph in the each region.

(a) In the left hand region,

f(-4) = (16-4)/(16-9) = 12/7 ≈ 1.7.

The point at ($- 4 , 1.7$) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) In the right hand region,

f(4) = (16-4)/(16-9) = 12/7 ≈ 1.7.

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

(c) In the middle region, we have

f(0) = 4/9 ≈ 0.44 and

f(-1) = f(1) = (1-4)/(1-9) =(-3)/-8 ≈ 0.38

The points at ($- 1 , 0.38$) and ($1 , 0.38$) are below the $y$-intercept, so we have an "inverted parabola" between the vertical asymptotes.

And we have our graph.