What is the vertices y=2x^2+12x+14 ?

The vertices y=2x^2+12x+14 is Minimum (-3,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=2x^2+12x+14

vertices

y = 2 x^{2} + 12 x + 14

Minimum (- 3, - 4)

Solution steps

y = 2 x^{2} + 12 x + 14

The parabola parameters are:

a = 2, b = 12, c = 14

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

x_{v} = - \frac{12}{2 \cdot 2}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 3, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 3, - 4)

x^{2} - 3 x - 4

4 x^{2} - 8 x - 3 y - 2 = 0

y = - \frac{( x + 3 ) ^{2}}{20} + 1

f (x) = 2 x^{2} + 4 x - 2

f (x) = 2 x^{2} + 4 x - 3

What is the vertices y=2x^2+12x+14 ?
The vertices y=2x^2+12x+14 is Minimum (-3,-4)