What is the vertices y=2x^2+4x-8 ?

The vertices y=2x^2+4x-8 is Minimum (-1,-10)

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vertices y=2x^2+4x-8

vertices

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

Minimum (- 1, - 10)

Solution steps

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

The parabola parameters are:

a = 2, b = 4, c = - 8

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

x_{v} = - \frac{4}{2 \cdot 2}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 10

Therefore the parabola vertex is

(- 1, - 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 10)

f (x) = - 2 x^{2} + 8 x - 11

f (x) = - 3 x^{2} + 4 x + 4

f (x) = 2 (x - 5) (x - 1)

- x^{2} + 4 x - 6

y = \frac{6 x}{7} \cdot \frac{4 x}{3}