What is the vertices y=-x^2+4x+12 ?

The vertices y=-x^2+4x+12 is Maximum (2,16)

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vertices y=-x^2+4x+12

vertices

y = - x^{2} + 4 x + 12

Maximum (2, 16)

Solution steps

y = - x^{2} + 4 x + 12

The parabola parameters are:

a = - 1, b = 4, c = 12

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(2, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 16)

f (x) = - 2 x^{2} - 20 x - 46

x^{2} - 4 x + 8

y = 3 x^{2} + 30 x + 65

f (x) = - x^{2} + 3 x + 4

y = (x - 8)^{2}