Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 36

Determine the horizontal and vertical asymptotes of the curve $\displaystyle y = \frac{1 + x^4}{x^2 - x^4}$

Solving for the vertical asymptotes

We set the denominator equal to zero


$
\begin{equation}
\begin{aligned}

x^2 - x^4 =& 0
\\
\\
x^2(1 - x^2) =& 0
\\
\\
x^2(1 - x)(1 + x) =&0

\end{aligned}
\end{equation}
$


$
\begin{array}{ccc}
x^2 = 0 & 1 - x = 0 & 1 + x = 0 \\
\sqrt{x^2} = \sqrt{0} & -x = -1 & x = -1 \\
\displaystyle x = 0 & \displaystyle \frac{-x}{-1} = \frac{-1}{-1} & \\
& x = 1 &
\end{array}
$

So the vertical asymptotes are $x = 0, x = 1$ and $x = -1$

Solving for the horizontal asymptotes

In the equation $\displaystyle y = \frac{1 + x^4}{x^2 - x^4}$ we remove everything except the biggest exponents of $x$ found in the numerator and denominator.

So we have


$
\begin{equation}
\begin{aligned}

y =& \frac{\cancel{x^4}}{-\cancel{x^4}}
\\
\\
y =& \frac{1}{-1}
\\
\\
y=& -1

\end{aligned}
\end{equation}
$



Thus, the horizontal asymptote is $y = -1$

Therefore,

the vertical asymptotes are $x = 0, x = 1$ and $x = -1$ and the horizontal asymptote is $y = -1$