What is the vertices y=x^2-10x+16 ?

The vertices y=x^2-10x+16 is Minimum (5,-9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices y=x^2-10x+16

vertices

y = x^{2} - 10 x + 16

Minimum (5, - 9)

Solution steps

y = x^{2} - 10 x + 16

The parabola parameters are:

a = 1, b = - 10, c = 16

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

x_{v} = - \frac{( - 10 )}{2 \cdot 1}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(5, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 9)