What is the vertices f(x)=-x^2-4x+12 ?

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

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vertices f(x)=-x^2-4x+12

vertices

f (x) = - 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

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} - 12 x + 10

18 x - 9 x^{2} - 9

+ s^{2} + 2 s + 2

f (x) = x^{2} + 7 x + 12

f (x) = x^{2} + 7 x + 10

What is the vertices f(x)=-x^2-4x+12 ?
The vertices f(x)=-x^2-4x+12 is Maximum (-2,16)