$\displaystyle y=3x-\frac{x^2}{2}$

Find y max

Here is how I'm approaching this ...

$\displaystyle 6x-x^2$

$\displaystyle y'=6-2x$

With 3 being a critical point

$\displaystyle y''=-2$

So the critical point is a maximum. It's a parabola opening downward from 3?