Question

What is the location of the vertex of the parabola defined by f(x) = −x2 − 8x − 11 and in what direction does the parabola open?

Answer 1

The vertex of a parabola is where dy/dx = 0
So we can differentiate that to find a solution:
$\frac{d}{dx}(-x^2-8x-11)=-2x-8 \\ -2x-8=0 \\ -2x=8 \\ x=-4$

Substituting in x=-4 into the original equation we can determine the y coordinate:
y = -(-4)^2-8(-4)-11 = -16+32-11 = 5

So we now know the coordinates of the vertex are (-4, 5)
There are lots of ways to work out the concavity of the parabola (i.e. which way it opens), and by default you can work this out because a -x^2 graph always opens downwards, however to prove this you can differentiate it again to work out the rate of change of gradient:

$\frac{d2y}{dx^2}=-2$
This shows the rate of change of gradient to be negative, i.e. the gradient is getting continuously more negative, which proves the parabola opens downwards.
You can also do this by plugging numbers around the vertex into the equation or the differential to work out the gradient.

So to summarise, the vertex is at (-4,5) and opens downwards.