# What are the asymptotes of f(x)=-x/((x-2)(4x-3) ?

Jun 26, 2018

Vertical Asymptote at $x = 2 , \frac{3}{4}$

Horizontal Asymptote at $y = 0$

#### Explanation:

Vertical Asymptotes are found where a function is undefined.

The only way to make our function undefined is having 0 in the Denominator.

$\left(x - 2\right) \left(4 x - 3\right) = 0$ $x = 2 , \frac{3}{4}$

These are our Vertical Asymptotes.

To find Horizontal Asymptotes we must look at our function at $- \infty , \infty$

${\lim}_{x \to - \infty} - \frac{x}{\left(x - 2\right) \left(4 x - 3\right)}$

Substitution gives $\frac{\infty}{\infty}$

Use $\textcolor{b l u e}{\text{L'Hopital's Rule}}$

${\lim}_{x \to - \infty} \left(\textcolor{b l u e}{\frac{d}{\mathrm{dx}}}\right) \cdot - \frac{x}{\left(x - 2\right) \left(4 x - 3\right)} = - \frac{1}{8 x - 11}$

Substitution give $- \frac{1}{-} \infty \rightarrow \frac{1}{\infty}$ which goes to $0$

Now look at the other side

${\lim}_{x \to \infty} - \frac{x}{\left(x - 2\right) \left(4 x - 3\right)}$

Substitution gives $- \frac{\infty}{\infty}$

Use $\textcolor{b l u e}{\text{L'Hopital's Rule}}$

${\lim}_{x \to \infty} \left(\textcolor{b l u e}{\frac{d}{\mathrm{dx}}}\right) \cdot - \frac{x}{\left(x - 2\right) \left(4 x - 3\right)} = - \frac{1}{8 x - 11}$

Substitute

$- \frac{1}{\infty}$ which goes to $0$

Limits agree $\therefore$ Horizontal Asymptote at $y = 0$