My teacher wrote that $$y + \frac{D}{4a} =a*(x+\frac {b}{2a})^2$$ is the value of slope . I thought it could be instantaneous slope then since slope is of a line and not a parabola. Then , my question is how did he derive this as value of slope.
My teacher said that equate the slope to 0. that is $dy/dx=0$
$dy/dx= 2a(x+\frac{b}{2a})$
so $2a(x+\frac{b}{2a})=0$
so $x=-b/2a$
As @Raffaele pointed out, what your teacher is attempting to do is to write the quadratic equation in vertex form, or to "Complete the square". Now at the vextex $\left(-\frac{b}{2a},-\frac{D}{4a}\right)$ the slope is $0$. I believe the goal is to show that you can find the $x$ value of the turning point either by completing the square or by setting the derivative to $0$ and solving for $x$.