Check my work? Find an equation for a graph whose vert. asymptote is at x = 0 and slant asymptote at y = x-1

Question

jennychan12 · Answer

$$y = \frac{ x^4+1 }{ x(x^2+1) }$$

jennychan12 · Answer

is there a specific way to do this? cuz i just guessed and checked.

anonymous · Answer

your equation is asymptotic to x.. more to come

jennychan12 · Answer

how do u do find an equation with slant asymptote to y = x-1 ?
that's the main part i'm stuck on

anonymous · Answer

(x^4-x^3)/(x^3+x) is asymptotic to x-1

anonymous · Answer

(x^4-x^3)/(x^3+x) is asymptotic to x-1

jennychan12 · Answer

how'd u find that? or did you just guess and check?

anonymous · Answer

it comes from how the first term in the denominator goes into the 1st term of the numerator. x^3 goes in to x^4: x times. and x^3 goes into -x^3: -1 times

jennychan12 · Answer

oh ok thanks :D

anonymous · Answer

The +x in the denominator irrelevent to the slant asymptote, because it is so much smaller than x^3 as x -> infinity