What is the vertices f(x)=(x-2)^2+3 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=(x-2)^2+3

vertices

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

Minimum (2, 3)

Solution steps

y = (x - 2)^{2} + 3

Rewrite

y = (x - 2)^{2} + 3

in the form

y = a x^{2} + b x + c

y = x^{2} - 4 x + 7

The parabola parameters are:

a = 1, b = - 4, c = 7

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 3)

y = 5 x^{2} + 13 x - 6

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

x^{2} - 12 x + 44

2 y^{2} - 8 x - 24 y + 56 = 0

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

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