Question

Find the horizontal and vertical asymptotes of the curve.
y = (1 + x^4) divided by (x^2 − x^4)

There are two y's and two x's. Thanks

Answer 1

vertical asymptotes are the points when the expression is undefined, the denominator is zero, the horizontal asymptotes are the found by the end behavior, when x approaches infinity, with huge values

Answer 2

first factor the numerator and denominator to make it easier

Answer 3

Can you explain? I don't know where to begin.

Answer 4

\(\bf y=\cfrac{1 + x^4}{{\color{red}{ x^2- x^4}}}\qquad \textit{horizontal asymptotes at }{\color{red}{ x^2- x^4}}=0\)

Answer 5

ahemm vertical rather

Answer 6

\(\bf y=\cfrac{1 + x^4}{{\color{red}{ x^2- x^4}}}\qquad \textit{vertical asymptotes at }{\color{red}{ x^2- x^4}}=0\)

Answer 7

so, solving that for "0", will give you values for "x", at those points is where the vertical ones are at

Answer 8

as far as the horizontal ones, depends on the degree of the numerator and denominator
to make it short
when both have the same degree, that is \(\bf y=\cfrac{1 + x^{\color{red}{ 4}}}{x^2- x^{\color{red}{ 4}}}\) then the horizontal asysmptote will be at the fraction made by coefficients of those leading terms, that is \(\bf y=\cfrac{1 + x^{\color{red}{ 4}}}{x^2- x^{\color{red}{ 4}}}\implies y=\cfrac{1 {\color{blue}{ +1}}x^{\color{red}{ 4}}}{x^2 {\color{blue}{ -1}}x^{\color{red}{ 4}}}\implies \cfrac{{\color{blue}{ +1}}}{{\color{blue}{ -1}}}\)

Answer 9

Thanks