I am stuck on graphing
$y= (x-3) \sqrt{x}$
I am pretty sure that the domain is all positive numbers including zero. The y intercept is 0, x is 0 and 3.
There are no asymptotes or symmetry.
Finding the interval of increase or decrease I take the derivative which will give me
$\sqrt{x} + \frac{x-3}{2\sqrt{x}}$
Finding zeroes for this I subtract $\sqrt{x}$ and then multiply by the denominator
$2x = x - 3$
This gives me 3 as a zero, but this isn't correct according to the book so I am stuck. I am not sure what is wrong.
Your solution of $y'=0$ is incorrect. $$\begin{align*} \sqrt{x}+\frac{x-3}{2\,\sqrt{x}}&=0\\ \frac{x-3}{2\,\sqrt{x}}&=-\sqrt{x}\\ x-3&=-2\,x\\ x&=? \end{align*}$$