# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x) = (4x) / (x^2+1)?

Dec 4, 2016

The horizontal asymptote is $y = 0$ and the function has neither a vertical asymptote nor an oblique asymptote.

#### Explanation:

$f \left(x\right) = \frac{4 x}{{x}^{2} + 1}$

To find the vertical asymptote (VA), find the value of $x$ that will make the denominator equal zero.

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

There is no real solution, so there is no VA.

To find the horizontal asymptote(HA), compare the degree of the numerator to the degree of the denominator.

$f \left(x\right) = \frac{4 {x}^{\textcolor{red}{1}}}{{x}^{\textcolor{b l u e}{2}} + 1}$

The degree of the numerator is $\textcolor{red}{1}$ and the degree of the denominator is $\textcolor{b l u e}{2}$.

If the degree of the numerator is less than the degree of the denominator, the HA is $y = 0$.

If the degree of the numerator is equal to the degree of the denominator, the HA is $y =$ the leading coefficient of the numerator divided by the leading coefficient of the denominator.

If the degree of the numerator is greater than the degree of the denominator, there is an oblique asymptote.

In this example, the degree of the numerator is less than the degree of the denominator, so the HA is $y = 0$ and there is no oblique asymptote.