What is the vertices f(x)=-4x^2-12x+5 ?

The vertices f(x)=-4x^2-12x+5 is Maximum (-3/2 ,14)

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vertices f(x)=-4x^2-12x+5

vertices

f (x) = - 4 x^{2} - 12 x + 5

Maximum (- \frac{3}{2}, 14)

Solution steps

y = - 4 x^{2} - 12 x + 5

The parabola parameters are:

a = - 4, b = - 12, c = 5

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} = 14

Therefore the parabola vertex is

(- \frac{3}{2}, 14)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (- \frac{3}{2}, 14)

What is the vertices f(x)=-4x^2-12x+5 ?
The vertices f(x)=-4x^2-12x+5 is Maximum (-3/2 ,14)