What is the vertices y=-2x^2-8x+3 ?

The vertices y=-2x^2-8x+3 is Maximum (-2,11)

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vertices y=-2x^2-8x+3

vertices

y = - 2 x^{2} - 8 x + 3

Maximum (- 2, 11)

Solution steps

y = - 2 x^{2} - 8 x + 3

The parabola parameters are:

a = - 2, b = - 8, c = 3

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

x_{v} = - \frac{( - 8 )}{2 ( - 2 )}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 11

Therefore the parabola vertex is

(- 2, 11)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- 2, 11)

x^{2} - 2 x + 1

y^{2} = 5 x + 3

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

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

f (x) = x^{2} + 12 x + 16

What is the vertices y=-2x^2-8x+3 ?
The vertices y=-2x^2-8x+3 is Maximum (-2,11)