What is the vertices-(x+2)^2+6 ?

The vertices-(x+2)^2+6 is Maximum (-2,6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-(x+2)^2+6

vertices

- (x + 2)^{2} + 6

Maximum (- 2, 6)

Solution steps

y = - (x + 2)^{2} + 6

Rewrite

y = - (x + 2)^{2} + 6

in the form

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

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

The parabola parameters are:

a = - 1, b = - 4, c = 2

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

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

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(- 2, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 6)

f (x) = 2 x^{2} + 3 x - 5

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

4 x - y^{2} + y - \frac{31}{4} = 0

f (x) = x^{2} - 10 x + 16

- 3 x^{2} - 5