Math
posted by CC .
Find the vertex of the parabola.
y = 4x2  16x  11

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~ :) 
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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