What is the vertices y=-x^2+2x-10 ?

The vertices y=-x^2+2x-10 is Maximum (1,-9)

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vertices y=-x^2+2x-10

vertices

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

Maximum (1, - 9)

Solution steps

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

The parabola parameters are:

a = - 1, b = 2, c = - 10

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

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

Simplify

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

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(1, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, - 9)

1 - x - x^{2}

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

x^{2} + 4

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

y^{2} - 2 y - 8 x - 31 = 0