Math Problem Solver

$f(x)=x^{2} + 6 - \frac{36}{3 x}$ has a vertical asymptote at $x=0$.

A line $x = a$ is a vertical asymptote of a function $f(x)$ if $\displaystyle \lim_{x \to a^+} f(x) = \pm \infty$ or $\displaystyle \lim_{x \to a^-} f(x) = \pm \infty$. To find vertical asymptotes of a function, search for points at which the function is undefined.

$f(x)=x^{2} + 6 - \frac{36}{3 x}$ contains $\frac{1}{3 x}$, so $f(x)=x^{2} + 6 - \frac{36}{3 x}$ is undefined where $3 x = 0$. Therefore, the function has a potential vertical asymptote at $x = 0$.

Take the limit of the function as $x$ approaches $0$ from both sides. $$\begin{aligned} \lim_{x \to 0^+}\left(x^{2} + 6 - \frac{36}{3 x}\right) &= -\infty\\ \lim_{x \to 0^-}\left(x^{2} + 6 - \frac{36}{3 x}\right) &= -\infty \end{aligned}$$The limit is infinite as $x$ approaches from either side, so $f(x)=x^{2} + 6 - \frac{36}{3 x}$ has a vertical asymptote at $x = 0$.