What is the vertices f(x)=4x^2+8x-8 ?

The vertices f(x)=4x^2+8x-8 is Minimum (-1,-12)

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vertices f(x)=4x^2+8x-8

vertices

f (x) = 4 x^{2} + 8 x - 8

Minimum (- 1, - 12)

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{8}{2 \cdot 4}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 12

Therefore the parabola vertex is

(- 1, - 12)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- 1, - 12)

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

f (x) = 4 x^{2} + 8 x + 3

x = 7 y^{2}

y = - x^{2} + 10 x - 28

+ y^{2} = 28 x

What is the vertices f(x)=4x^2+8x-8 ?
The vertices f(x)=4x^2+8x-8 is Minimum (-1,-12)