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

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

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vertices y=-4x^2-12x-8

vertices

y = - 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} + 8 x + 9

25 x^{2} - 36

y = 4 x^{2} - 40 x + 98

x = 2 y^{2}

x^{2} - 14 x + 49

What is the vertices y=-4x^2-12x-8 ?
The vertices y=-4x^2-12x-8 is Maximum (-3/2 ,1)