What is the vertices y=-x^2-4x+1 ?

The vertices y=-x^2-4x+1 is Maximum (-2,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=-x^2-4x+1

vertices

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

Maximum (- 2, 5)

Solution steps

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

The parabola parameters are:

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

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} = 5

Therefore the parabola vertex is

(- 2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 5)

x^{2} - 3 x + 5

3 x^{2} - 3 x

y = 2 x^{2} + 12 x + 17

4 x^{2} + 8 x + 4

y = (x + 2)^{2} + 1