What is the vertices g(x)=(x+3)^2+5 ?

The vertices g(x)=(x+3)^2+5 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 g(x)=(x+3)^2+5

vertices

g (x) = (x + 3)^{2} + 5

Minimum (- 3, 5)

Solution steps

y = (x + 3)^{2} + 5

Rewrite

y = (x + 3)^{2} + 5

in the form

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

y = x^{2} + 6 x + 14

The parabola parameters are:

a = 1, b = 6, c = 14

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

x_{v} = - \frac{6}{2 \cdot 1}

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 = 1

Minimum (- 3, 5)

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

f (x) = x^{2} + 8 + 15

y = x^{2} + 4 x - 5

3 x^{2} + 1

y = 7 x^{2} + 5

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