What is the vertices 5x^2+10x+6 ?

The vertices 5x^2+10x+6 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 5x^2+10x+6

vertices

5 x^{2} + 10 x + 6

Minimum (- 1, 1)

Solution steps

y = 5 x^{2} + 10 x + 6

The parabola parameters are:

a = 5, b = 10, c = 6

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

x_{v} = - \frac{10}{2 \cdot 5}

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 = 5

Minimum (- 1, 1)

+ x^{2} - 5

x^{2} + 8 x + 19

y = x^{2} + 7 x + 12

y = x^{2} - 12 x - 20

f (x) = - 2 x^{2} - x + 1