Find the Asymptotes y=(x^2+6)/(16x^2-3)

Find where the expression ${\displaystyle \frac{x^2+6}{16x^2-3}}$ is undefined.

Since ${\displaystyle \frac{x^2+6}{16x^2-3}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -\frac{\sqrt{3}}{4}}$ from the left and ${\displaystyle \frac{x^2+6}{16x^2-3}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle -\frac{\sqrt{3}}{4}}$ from the right, then ${\displaystyle x=-\frac{\sqrt{3}}{4}}$ is a vertical asymptote.

Since ${\displaystyle \frac{x^2+6}{16x^2-3}}$${\displaystyle \to }$${\displaystyle -\infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle \frac{\sqrt{3}}{4}}$ from the left and ${\displaystyle \frac{x^2+6}{16x^2-3}}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle \frac{\sqrt{3}}{4}}$ from the right, then ${\displaystyle x=\frac{\sqrt{3}}{4}}$ is a vertical asymptote.

List all of the vertical asymptotes:

Consider the rational function ${\displaystyle R\left(x\right)=\frac{a{x}^n}{b{x}^m}}$ where ${\displaystyle n}$ is the degree of the numerator and ${\displaystyle m}$ is the degree of the denominator.

Find ${\displaystyle n}$ and ${\displaystyle m}$.

Since ${\displaystyle n=m}$, the horizontal asymptote is the line ${\displaystyle y=\frac{a}{b}}$ where ${\displaystyle a=1}$ and ${\displaystyle b=16}$.

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

This is the set of all asymptotes.