What is the vertices-4x^2-12x-8 ?

The vertices-4x^2-12x-8 is Maximum (-3/2 ,1)

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vertices-4x^2-12x-8

vertices

- 4 x^{2} - 12 x - 8

Maximum (- \frac{3}{2}, 1)

Solution steps

y = - 4 x^{2} - 12 x - 8

The parabola parameters are:

a = - 4, b = - 12, c = - 8

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

x_{v} = - \frac{( - 12 )}{2 ( - 4 )}

Simplify

- \frac{- 12}{2 ( - 4 )} : - \frac{3}{2}

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- \frac{3}{2}, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (- \frac{3}{2}, 1)

y = x^{2} - 5 x - 14

x^{2} + 14 x - 8 y + 65 = 0

x^{2} - 4 x - 4

f (x) = 2 x^{2} - 6 x + 15

f (x) = 2 x^{2} + 5 x - 3