# How do you find the asymptotes for f(x)= (4x^2+5)/( x^2-1)?

Mar 11, 2016

To find vertical asymptote, you must set the denominator to 0 and then solve.

#### Explanation:

${x}^{2} - 1 = 0$

This can be factored as a difference of squares $\left({a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)\right)$

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

$x = - 1 \mathmr{and} 1 \implies$ there are vertical asymptotes at x = -1 and x = 1.

Now for horizontal asymptotes, which are a little trickier.

To find these, you must look for the highest power (exponent) in both the numerator and the denominator. The highest power in both is ${x}^{2}$, which means that the asymptote will occur at the ratio of the coefficients of the highest power in the numerator and denominator. The coefficients are 4 and 1 respectively.

So, the horizontal asymptote occurs at $y = 4$

Verification:

Plugging in 4 for $f \left(x\right)$

$4 = \frac{4 {x}^{2} + 5}{{x}^{2} - 1}$

$4 \left({x}^{2} - 1\right) = 4 {x}^{2} + 5$

$4 {x}^{2} - 4 = 4 {x}^{2} + 5$

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

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

$x = \emptyset$, since division by 0 is undefined.

This proves that there is a horizontal asymptote at y = 4, because an asymptote is essentially and undefined line on the graph of a function.

Practice exercises:

1. Identify all the asymptotes in $g \left(x\right) = \frac{5 {x}^{2} + 10 x}{2 {x}^{2} + 7 x + 3}$

Good luck!