Question

find the horizontal and vertical asymptotes of the curve. you may want to use a graphing calculator (or computer) to check your work by graphing the curve and estimating the asymptotes.
(enter your answers as comma - separated lists. if an answer does not exist, enter dne.)

y=\frac{6+x^{4}}{x^{2}-x^{4}}

( x= )

( y= )

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero).
Set \(x^{2}-x^{4}=x^{2}(1 - x^{2})=x^{2}(1 - x)(1 + x)=0\).
Solving \(x^{2}(1 - x)(1 + x)=0\), we get \(x = 0,x = 1,x=-1\).
Check numerator at these values:
When \(x = 0\), \(y=\frac{6+0}{0 - 0}\) (undefined, but since \(x^{2}\) is a factor of denominator and not of numerator, \(x = 0\) is not an asymptote).
When \(x=1\), \(y=\frac{6 + 1}{1-1}\) (undefined and numerator \(6 + 1=7
eq0\)).
When \(x=-1\), \(y=\frac{6+1}{1 - 1}\) (undefined and numerator \(6 + 1=7
eq0\)).

Step2: Find horizontal asymptotes

For horizontal asymptotes, use the limit as \(x
ightarrow\pm\infty\).
Divide numerator and denominator by \(x^{4}\):
\(y=\lim_{x
ightarrow\pm\infty}\frac{\frac{6}{x^{4}}+1}{\frac{1}{x^{2}}-1}\)
As \(x
ightarrow\pm\infty\), \(\lim_{x
ightarrow\pm\infty}\frac{6}{x^{4}} = 0\) and \(\lim_{x
ightarrow\pm\infty}\frac{1}{x^{2}}=0\)
So \(y=\frac{0 + 1}{0-1}=-1\)

Answer:

\(x = 1,-1\)
\(y=-1\)