What is the vertices f(x)=-x^2-6x+3 ?

The vertices f(x)=-x^2-6x+3 is Maximum (-3,12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=-x^2-6x+3

vertices

f (x) = - x^{2} - 6 x + 3

Maximum (- 3, 12)

Solution steps

y = - x^{2} - 6 x + 3

The parabola parameters are:

a = - 1, b = - 6, c = 3

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

x_{v} = - \frac{( - 6 )}{2 ( - 1 )}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 12

Therefore the parabola vertex is

(- 3, 12)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 3, 12)

x^{2} - 2 x - 3

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

(x + 5)^{2}

y = 4 x^{2} - 4 x^{1}

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

What is the vertices f(x)=-x^2-6x+3 ?
The vertices f(x)=-x^2-6x+3 is Maximum (-3,12)