What is the vertices f(x)=-2x^2-4x+13 ?

The vertices f(x)=-2x^2-4x+13 is Maximum (-1,15)

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vertices f(x)=-2x^2-4x+13

vertices

f (x) = - 2 x^{2} - 4 x + 13

Maximum (- 1, 15)

Solution steps

y = - 2 x^{2} - 4 x + 13

The parabola parameters are:

a = - 2, b = - 4, c = 13

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 15

Therefore the parabola vertex is

(- 1, 15)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 1, 15)

y^{2} = - 16 x

y = x^{2} - 6 x - 1

y = x^{2} - 5 x - 24

2 x^{2} + 5 x - 12

f (x) = - x^{2} - 4 x + 12

What is the vertices f(x)=-2x^2-4x+13 ?
The vertices f(x)=-2x^2-4x+13 is Maximum (-1,15)