What is the vertices y=x^2-6x ?

The vertices y=x^2-6x is Minimum (3,-9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=x^2-6x

vertices

y = x^{2} - 6 x

Minimum (3, - 9)

Solution steps

y = x^{2} - 6 x

The parabola parameters are:

a = 1, b = - 6, c = 0

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

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

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(3, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 9)

f (x) = \frac{3 - 2 x}{2 - 3 x}

f (x) = - (x^{2})

\frac{3 x + 2}{x + 2}

f (x) = x^{2} - 10 x

f (x) = 3 x^{2} + 10 x - 2