What is the vertices f(x)=x^2-6x+4 ?

The vertices f(x)=x^2-6x+4 is Minimum (3,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2-6x+4

vertices

f (x) = x^{2} - 6 x + 4

Minimum (3, - 5)

Solution steps

y = x^{2} - 6 x + 4

The parabola parameters are:

a = 1, b = - 6, c = 4

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} = - 5

Therefore the parabola vertex is

(3, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 5)

x^{2} + x - 4

x^{2} + x - 6

f (x) = 0.75 x^{2} - 5

x^{2} + 9 x + 4

5 x^{2} + 10 x + 6

What is the vertices f(x)=x^2-6x+4 ?
The vertices f(x)=x^2-6x+4 is Minimum (3,-5)