Posted by **CC** on Tuesday, June 7, 2011 at 10:51pm.

Find the vertex of the parabola.

y = -4x2 - 16x - 11

- Math -
**Jai**, Tuesday, June 7, 2011 at 11:07pm
y = -4x^2 - 16x - 11

there are many ways to determine the vertex, but the easiest (for me) i think is to use derivatives. first we get the derivative with respect to x:

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

y' = -8x - 16

then we equate this to zero (because vertex is a maximum or a minimum and therefore the slope is zero):

0 = -8x - 16

8x = -16

x = -2

substitute this back to the original given equation:

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

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

y = -16 + 32 -11

y = 5

therefore vertex is at

(-2, 5)

hope this helps~ :)

- or - Math -
**Reiny**, Tuesday, June 7, 2011 at 11:26pm
If you don't know Calculus, like in Jai's method,

then....

the x value of the vertex is -b/(2a) = 16/-8 = -2

sub into original to get

y = -16 + 32 - 11 = 5

vertex is (-2,5)

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