What is the vertices f(x)=2x^2-12x+23 ?

The vertices f(x)=2x^2-12x+23 is Minimum (3,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=2x^2-12x+23

vertices

f (x) = 2 x^{2} - 12 x + 23

Minimum (3, 5)

Solution steps

y = 2 x^{2} - 12 x + 23

The parabola parameters are:

a = 2, b = - 12, c = 23

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} = 5

Therefore the parabola vertex is

(3, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, 5)

x^{2} + 6 x + 7

f (x) = 6 x^{2} - x + 1

y = 2 x^{2} + 10 x + 12

x^{2} + 4 x - 21

x^{2} + 4 x - 12

What is the vertices f(x)=2x^2-12x+23 ?
The vertices f(x)=2x^2-12x+23 is Minimum (3,5)