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~ :)
Find the vertex of the parabola.
y = -4x2 - 16x - 11
2 answers
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)
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)