What is the vertices-x^2-4x+5 ?

The vertices-x^2-4x+5 is Maximum (-2,9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-x^2-4x+5

vertices

- x^{2} - 4 x + 5

Maximum (- 2, 9)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 4, c = 5

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} = 9

Therefore the parabola vertex is

(- 2, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 9)

f (x) = x^{2} + 16 x + 64

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

y = 6 x^{2} - 1

(x + 5)^{2} = - 20 (y - 2)

7 x^{2} - 10 x + 5