What is the vertices y=-2x^2-4x+6 ?

The vertices y=-2x^2-4x+6 is Maximum (-1,8)

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vertices y=-2x^2-4x+6

vertices

y = - 2 x^{2} - 4 x + 6

Maximum (- 1, 8)

Solution steps

y = - 2 x^{2} - 4 x + 6

The parabola parameters are:

a = - 2, b = - 4, c = 6

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} = 8

Therefore the parabola vertex is

(- 1, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 1, 8)

f (x) = x^{2} - 6

x^{2} + 5 x - 1

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

- 3 x^{2} + 2 x + 1

y = (x - 3)^{2} + 2