Find the vertex of the parabola y -4x2 - 16x - 11?

Answer 1

Start by finding the x-coordinate of that vertex. You can do that by taking the derivative of the function and solving for zero:

y = -4x2 - 16x - 11

y' = -8x - 16

0 = -8x - 16

8x = -16

x = -2

Now simply plug the x-coordinate into the original equation to get your y-coordinate:

y = -4(-2)2 - 16(-2) - 11

y = -16 + 32 - 11

y = 5

So the vertex occurs at the point (-2, 5)

Alternative answer:

It should be noted that the above method requires a little extra work if you are not working with a parabola, as the first derivative only allows you to find the critical points of a function. With an arbitrary function, you also need to take the second derivative of the function to determine if the critical point is a maximum, minimum or point of inflection.

If you don't know Calculus yet, there is also an algebraic method to find the vertex of a parabola

We have:

y = -4x^2 - 16x - 11

This is the parabola's standard form. In order to find the vertex through algebra we need to convert it to vertex form:

y = a(x - h) + k

Where (h,k) is our vertex

We can do this by factoring (which is always a pain):

First factor out a -4:

y = -4(x^2 + 4x) - 11

We left the -11 alone because 11/4 is pretty ugly to work with, so we will leave it on the side for now.

Next we complete the square:

y = -4(x^2 + 4x + c - c) - 11

y = -4(x^2 + 4x + c) - 11 + 4c

We want to find c, such that 2*(c^(1/2)) = 4, which gives us c = 4.

y = -4(x^2 + 4x + 4) - 11 + 16

y = -4(x + 2)^2 + 5

Comparing with our vertex form:

y = a(x - h) + k

We have a = -4, h = -2 and k = 5.

(h, k) is our vertex, which gives us (-2, 5). This is consistent with the answer given by the Calculus method above, which is reassuring.