# How do you graph (4x+1)/(x^2-1 using asymptotes, intercepts, end behavior?

Nov 9, 2017

See below.

#### Explanation:

$\frac{4 x + 1}{{x}^{2} - 1}$ is undefined for $x = 1 \mathmr{and} - 1$ ( Division by zero )

So the line $x = 1 \mathmr{and} x = - 1$ are vertical asymptotes.

Y axis intercept occurs when $x = 0$

$y = \frac{4 \left(0\right) + 1}{{\left(0\right)}^{2} - 1} = - 1$ ( 0 , -1 )

x axis intercepts occur when y = 0.

$\frac{4 x + 1}{{x}^{2} - 1} = 0$

We can solve this by reasoning that if $\left(4 x + 1\right) = 0$

Then:

$\frac{4 x + 1}{{x}^{2} - 1} = 0$

$\therefore$

$\left(4 x + 1\right) = 0 \implies x = - \frac{1}{4}$

Note: Do not try to solve for $\left({x}^{2} - 1\right) = 0$ ( zero denominator is invalid )

x axis intercept at:

$\left(- \frac{1}{4} , 0\right)$

For limits to infinity we only need to concern ourselves with the two terms containing the variable.

$\frac{4 x}{x} ^ 2$

$\frac{4 x}{x} ^ 2 = \frac{4}{x}$

${\lim}_{x \to \infty} \left(\frac{4}{x}\right) = 0$

${\lim}_{x \to - \infty} \left(\frac{4}{x}\right) = 0$

So the x axis is a horizontal asymptote.

Graph: