Question

17x/(x^4 + 1)^1/4...what are the horizontal asymptotes.

only the denominator is raised to 1/4

Answer 1

Since the lead exponents are equal, the horizontal asymptote is the ratio of the coefficients of the leading terms.

Answer 2

what mertsj said, with the sophistication that as x gets large
\[\sqrt[4]{x^4+1}\] behaves like
\[\sqrt[4]{x}=x\] so you might as well pretend you have
\[\frac{17x}{x}=17\]

Answer 3

oh i see. how do i find y2

Answer 4

What do you mean, y2?

Answer 5

y1= and y2= where y1>y2

Answer 6

And don't forget there are two fourth roots of one so there are two horizontal asymptotes, y = 17 and y=-17

Answer 7

thats what i meant, thanks

Answer 8

\[y=\frac{17x}{(x^4+1)^{\frac{1}{4}}}\]
Rearrange to get x in term of y:
\[y(x^4+1)^{\frac{1}{4}} =17x\]
\[y^4(x^4+1) = 83521x^4\]
\[\frac{(x^4+1)}{x^4} = \frac{83521}{y^4}\]
\[\frac{1}{x^4} = \frac{83521}{y^4}-1=\frac{83521 - y^4}{y^4}\]
So \[x =\pm \left(\frac{y^4}{83521 - y^4}\right)^{\frac{1}{4}}\]
So the value of y where the denominator is zero is the value at which the horizontal asymtote occurs...
\[y^4 = 83521\] so \[y = \pm 17\]

Answer 9

that's how you prove it the long way round I think...