## anonymous 4 years ago 17x/(x^4 + 1)^1/4...what are the horizontal asymptotes. only the denominator is raised to 1/4

1. Mertsj

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

2. anonymous

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 $\frac{17x}{x}=17$

3. anonymous

oh i see. how do i find y2

4. Mertsj

What do you mean, y2?

5. anonymous

y1= and y2= where y1>y2

6. Mertsj

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

7. anonymous

thats what i meant, thanks

8. anonymous

$y=\frac{17x}{(x^4+1)^{\frac{1}{4}}}$ 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 $y = \pm 17$

9. anonymous

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