What is the vertices f(x)=-2x^2+80x-10 ?

The vertices f(x)=-2x^2+80x-10 is Maximum (20,790)

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vertices f(x)=-2x^2+80x-10

vertices

f (x) = - 2 x^{2} + 80 x - 10

Maximum (20, 790)

Solution steps

y = - 2 x^{2} + 80 x - 10

The parabola parameters are:

a = - 2, b = 80, c = - 10

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

x_{v} = - \frac{80}{2 ( - 2 )}

Simplify

- \frac{80}{2 ( - 2 )} : 20

x_{v} = 20

Plug in

x_{v} = 20

to find the

y_{v}

value

y_{v} = 790

Therefore the parabola vertex is

(20, 790)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (20, 790)

y = 6 x^{2} - 7 x - 3

f (x) = 4 + 12 x - 3 x^{2}

f (x) = - 5 x^{2} - 25 x - 30

y = x^{2} - 6 x + 9

y = 2 x^{2} - 13 x - 7

What is the vertices f(x)=-2x^2+80x-10 ?
The vertices f(x)=-2x^2+80x-10 is Maximum (20,790)