What is the vertices f(x)=3+x-2x^2 ?

The vertices f(x)=3+x-2x^2 is Maximum (1/4 , 25/8)

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vertices f(x)=3+x-2x^2

vertices

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

Maximum (\frac{1}{4}, \frac{25}{8})

Solution steps

y = 3 + x - 2 x^{2}

Rewrite

y = 3 + x - 2 x^{2}

in the form

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

y = - 2 x^{2} + x + 3

The parabola parameters are:

a = - 2, b = 1, c = 3

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

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

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{25}{8}

Therefore the parabola vertex is

(\frac{1}{4}, \frac{25}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (\frac{1}{4}, \frac{25}{8})

y = 2 x^{2} - 6 x + 4

y = x^{2} - 16

y = x^{2} - 13

f (x) = 3 x^{2} - 2 x

y = - x^{2} + 80 x - 700

What is the vertices f(x)=3+x-2x^2 ?
The vertices f(x)=3+x-2x^2 is Maximum (1/4 , 25/8)