What is the vertices 2x^2+4x+1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2x^2+4x+1

vertices

2 x^{2} + 4 x + 1

Minimum (- 1, - 1)

Solution steps

y = 2 x^{2} + 4 x + 1

The parabola parameters are:

a = 2, b = 4, c = 1

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

x_{v} = - \frac{4}{2 \cdot 2}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 1)

- x^{2} + 2 x + 8

y = x^{2} + 2 x - 1

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

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

x^{2} + 5 x - 1