fractional exponents

[panda411]

In my Calculus 1 class, we are working on curve sketching. My professor wants us to use long division to find asymptopes and then use first and second derivatives to sketch the curve. I understand the steps, I just cannot get a fraction of of the equation.
The equations are y=x^(1/3) * (x-4)
and y=x^(2/3) * (x-7)

I just cannot see how to get a fraction out of these equations. I see how (x^(1/3)) can become a cube root, but I cannot find the fraction. My professor said that equation would have a vertical asymptope at x=0. If I could somehow get the cube root in the demonator, I can see how that would be true.

Any help will be appreciated.
Thanks!

 

[soroban]

Hello, panda411!

The first one is: \(\displaystyle \L\,y \:=\:x^{\frac{1}{3}}(x\,-\,4)\)

Differentiate (product rule): \(\displaystyle \L\,y' \:=\:x^{\frac{1}{3}}\cdot1 \,+\,\frac{1}{3}x^{-\frac{2}{3}}(x\,-\,4) \:=\:x^{-\frac{1}{3}} \,+\,\frac{x\,-\,4}{3x^{\frac{2}{3}}}\)

Get a common denominator: \(\displaystyle \L\,y' \:=\:x^{\frac{1}{3}}\cdot\frac{3x^{\frac{2}{3}}}{3x^{\frac{2}{3}}}\,+\,\frac{x\,-\,4}{3x^{\frac{2}{3}}} \:=\:\frac{3x\,+\,(x\,-\,4)}{3x^{\frac{2}{3}}} \;=\;\frac{4x\,-\,4}{3x^{\frac{2}{3}}}\)

We have: \(\displaystyle \L\,y'\:=\:\frac{4(x\,-\,1)}{3x^{\frac{2}{3}}}\)

There is a horizontal tangent at \(\displaystyle x\,=\,1\)
. . and a vertical tangent (vertical asymptote) at \(\displaystyle x\,=\,0.\)