What is the vertices y=-x^2+80x-700 ?

The vertices y=-x^2+80x-700 is Maximum (40,900)

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vertices y=-x^2+80x-700

vertices

y = - x^{2} + 80 x - 700

Maximum (40, 900)

Solution steps

y = - x^{2} + 80 x - 700

The parabola parameters are:

a = - 1, b = 80, c = - 700

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

x_{v} = - \frac{80}{2 ( - 1 )}

Simplify

- \frac{80}{2 ( - 1 )} : 40

x_{v} = 40

Plug in

x_{v} = 40

to find the

y_{v}

value

y_{v} = 900

Therefore the parabola vertex is

(40, 900)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (40, 900)

x^{2} + 2 x - 24

y = 2 x^{2} + 12 x + 14

x^{2} - 3 x - 4

4 x^{2} - 8 x - 3 y - 2 = 0

y = - \frac{( x + 3 ) ^{2}}{20} + 1

What is the vertices y=-x^2+80x-700 ?
The vertices y=-x^2+80x-700 is Maximum (40,900)