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

The vertices 4x^2+4x-8 is Minimum (-1/2 ,-9)

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vertices 4x^2+4x-8

vertices

4 x^{2} + 4 x - 8

Minimum (- \frac{1}{2}, - 9)

Solution steps

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

The parabola parameters are:

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

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

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

Simplify

- \frac{4}{2 \cdot 4} : - \frac{1}{2}

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(- \frac{1}{2}, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- \frac{1}{2}, - 9)

f (x) = - 2 x^{2} - 16 x - 31

- x^{2} + 2 x + 2

f (x) = - 2 x^{2} - 16 x - 37

- x^{2} + 2 x + 4

4 x^{2} - 8 x - 2