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

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

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

vertices

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

Maximum (1, 1)

Solution steps

y = - x^{2} + 2 x

The parabola parameters are:

a = - 1, b = 2, c = 0

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} = 1

Therefore the parabola vertex is

(1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 1)

y = (2 x + 5)^{2}

f (x) = - 9 x^{2} + 9 x + 1

+ x^{2} + 4 x - 1

y^{2} - 12 y - 4 x = - 28

24000 x^{2} + 900 x - 150