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

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

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vertices-x^2+2x-10

vertices

- 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)

f (x) = 3 x^{2} - 12 x + 13

y = x^{2} - x - 6

f (x) = - x^{2} + 6 x - 8

- x^{2} + 5 x - 7

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