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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

vertices

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})

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