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