What is the vertices f(x)=(x-5)(x+1) ?

The vertices f(x)=(x-5)(x+1) is Minimum (2,-9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices f(x)=(x-5)(x+1)

vertices

Minimum (2, - 9)

Solution steps

y = (x - 5) (x + 1)

y = (x - 5) (x + 1)

The parabola parameters are:

a = 1, m = 5, n = - 1

x_{v} = \frac{m + n}{2}

x_{v} = \frac{5 + ( - 1 )}{2}

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

Minimum (2, - 9)

What is the vertices f(x)=(x-5)(x+1) ?
The vertices f(x)=(x-5)(x+1) is Minimum (2,-9)