# How do you sketch the curve for y= (x^2+1)/(x^2-4)?

Aug 2, 2015

Find the critical numbers, local maximum and determine the behaviors as $\pm \infty$.

#### Explanation:

Step 1. Determine the domain and range of $f \left(x\right) = \frac{{x}^{2} + 1}{{x}^{2} - 4}$.

Domain is all values of $x$ except where
${x}^{2} - 4 = 0$
${x}^{2} = 4$
$x = \pm 2$

Range is $\left(- \infty , \infty\right)$

Step 2. Find the critical numbers of $f \left(x\right)$ in $\left(- 2 , 2\right)$.
Apply the Quotient Rule to differentiate the function.
$f ' \left(x\right) = \frac{\left({x}^{2} - 4\right) \left(2 x\right) - \left({x}^{2} + 1\right) \left(2 x\right)}{{x}^{2} - 4} ^ 2$
$f ' \left(x\right) = \frac{2 x}{{x}^{2} - 4} - \frac{2 x \left({x}^{2} + 1\right)}{{x}^{2} - 4} ^ 2$

Critical numbers:
$x = \pm 2$, where $f ' \left(x\right)$ does not exist, and $x = 0$, where $f ' \left(x\right) = 0$.

Step 3. Calculate $f \left(x\right)$ with critical numbers.
$f \left(- 2\right)$ does not exist. The limit from the left is $+ \infty$ and the limit from the right is $- \infty$.
$f \left(2\right)$ does not exist. The limit from the left is $- \infty$ and the limit from the right is $+ \infty$.
$f \left(0\right) = - \frac{1}{4}$ is a local maximum.
For $x < - 2$, $f \left(x\right) > 0$. For $x > 2$, $f \left(x\right) > 0$.

Step 4. Sketch the graph with these attributes.

graph{(x^2+1)/(x^2-4) [-20, 20, -10, 10]}