How do you find vertical, horizontal and oblique asymptotes for #f(x) =(4x^5)/(x^3-1)#?

1 Answer
Dec 6, 2016

Answer:

Vertical+ #darr x = 1 uarr# The contracted graph illustrates the asymptoticity of x = 1.

Explanation:

By actual division #y = 4x^2 +(4x^2)/((x-1)(x^2+x+1))#.

As #x to 1, y to 4(1+-oo(1/3))=+-oo#.

So, x = 1 is the asymptote.

I surmise that the origin is not point of inflexion, despite that

seemingly it is so.

The curve is flat there, upon the x-axis, in infinitesimal

neighborhood. In other words, y and all its derivatives are 0, And so,

thereabout the origin , there is no series expansion.

graph{y(x^3-1)-4x^5=0 [-40, 40, -20, 20]} }