What is the vertices 4x-2x^2 ?

The vertices 4x-2x^2 is Maximum (1,2)

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vertices 4x-2x^2

vertices

4 x - 2 x^{2}

Maximum (1, 2)

Solution steps

y = 4 x - 2 x^{2}

Rewrite

y = 4 x - 2 x^{2}

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

a = - 2, b = 4, c = 0

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

x_{v} = - \frac{4}{2 ( - 2 )}

Simplify

- \frac{4}{2 ( - 2 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (1, 2)

y = \frac{95 x ^{2} - 10 x + 26}{10}

(x - 1)^{2} = - 2 y - 11

f (x) = x^{2} - 6 x + 10

- x^{2} + 4 x

f (x) = - 6 x^{2} + 17 x - 12