# How do you find all the asymptotes for function f(x)= (17x)/( 2x^2 + 3)?

Sep 24, 2015

Its horizontal asymptote is the $x$-axis ($y = 0$) as $x \to \pm \infty$ and it has no vertical asymptotes.

#### Explanation:

The line $y = 0$ is a horizontal asymptote of the rational function $f \left(x\right) = \frac{17 x}{2 {x}^{2} + 3}$ since the degree of the bottom (denominator) is greater than the degree of the top (numerator). Furthermore, the denominator $2 {x}^{2} + 3$ has no real roots (the roots are the imaginary numbers $\pm i \sqrt{\frac{3}{2}}$). Hence, the function has no vertical asymptotes.

To confirm the horizontal asymptote, you can do the following limit calculation, which I'm not rigorously justifying:

${\lim}_{x \to \pm \infty} \frac{17 x}{2 {x}^{2} + 3} = {\lim}_{x \to \pm \infty} \frac{17 x}{2 {x}^{2} + 3} \cdot \frac{\frac{1}{x} ^ 2}{\frac{1}{x} ^ 2}$

$= {\lim}_{x \to \pm \infty} \frac{\frac{17}{x}}{2 + \frac{3}{x} ^ 2} = \frac{{\lim}_{x \to \pm \infty} \left(\frac{17}{x}\right)}{{\lim}_{x \to \pm \infty} 2 + {\lim}_{x \to \pm \infty} \left(\frac{3}{x} ^ 2\right)} = \frac{0}{2 + 0} = 0$