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

The vertices f(x)=-x^2+x is Maximum (1/2 , 1/4)

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

vertices

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

Maximum (\frac{1}{2}, \frac{1}{4})

Solution steps

y = - x^{2} + x

The parabola parameters are:

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

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

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

Simplify

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = \frac{1}{4}

Therefore the parabola vertex is

(\frac{1}{2}, \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{1}{2}, \frac{1}{4})

(x + 2)^{2} - 9

y = x^{2} - 8 x + 7

f (x) = - x^{2} + 7

y = - 4 x^{2} + 8 x + 2

y^{2} + 12 x - 2 y - 11 = 0

What is the vertices f(x)=-x^2+x ?
The vertices f(x)=-x^2+x is Maximum (1/2 , 1/4)