What is the vertices y=-x^2+2x-1 ?

The vertices y=-x^2+2x-1 is Maximum (1,0)

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vertices y=-x^2+2x-1

vertices

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

Maximum (1, 0)

Solution steps

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

The parabola parameters are:

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

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} = 0

Therefore the parabola vertex is

(1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 0)

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

2 x^{2} - x

2 x^{2} - 8

y = - x^{2} + 4 x + 12

f (x) = - 2 x^{2} - 20 x - 46