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

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

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

vertices

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

Maximum (- 2, 2)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 4, c = - 2

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} = 2

Therefore the parabola vertex is

(- 2, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 2)

(x - 1)^{2} = - 4 (y - 2)

f (x) = \frac{1}{3} x^{2} + 2 x + 1

4 x^{2} - 8 x + 2 y = 5

y = 2 x^{2} + 5 x - 7

y = x (x - 18)

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