What is the vertices x^2+10x+11 ?

The vertices x^2+10x+11 is Minimum (-5,-14)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices x^2+10x+11

vertices

Minimum (- 5, - 14)

Solution steps

y = x^{2} + 10 x + 11

The parabola parameters are:

a = 1, b = 10, c = 11

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

x_{v} = - \frac{10}{2 \cdot 1}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = - 14

Therefore the parabola vertex is

(- 5, - 14)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 14)