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

The vertices 2x^2+2x-4 is Minimum (-1/2 ,-9/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2x^2+2x-4

vertices

2 x^{2} + 2 x - 4

Minimum (- \frac{1}{2}, - \frac{9}{2})

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{2}{2 \cdot 2}

Simplify

- \frac{2}{2 \cdot 2} : - \frac{1}{2}

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{9}{2}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{9}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{1}{2}, - \frac{9}{2})

x^{2} + 2 x + 3

y = - x^{2}

x^{2} - 4 x + 4

t^{2}

f (x) = - x^{2} - 6 x + 3