What is the vertices-5x^2-40x-77 ?

The vertices-5x^2-40x-77 is Maximum (-4,3)

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vertices-5x^2-40x-77

vertices

- 5 x^{2} - 40 x - 77

Maximum (- 4, 3)

Solution steps

y = - 5 x^{2} - 40 x - 77

The parabola parameters are:

a = - 5, b = - 40, c = - 77

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

x_{v} = - \frac{( - 40 )}{2 ( - 5 )}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 4, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (- 4, 3)

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

y = 5 x^{2} - 2 x - 1

f (x) = x^{2} - 10 x + 21

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

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