a) x^2 is always + so min when x = 0 of 12
b) if you know calculus
y' = -6x -12 = 0 at max Because -3x^2 forces parabola to open down)
x = -2
so y = -3(4)+24+15 = 27
If you do not know calculus, find the vertex by completing the square.
c) y = 3 x^2 -6 x + 5
y' = 6x-6 = 0 at min
x = 1 at min
y = 2
determine the maximum or the minimum value for each
a)y=2x^2+12
b)y=-3x^2-12x+15
c)3x(x-2)+5
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