What is the vertices f(x)=x^2-5x+4 ?

The vertices f(x)=x^2-5x+4 is Minimum (5/2 ,-9/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2-5x+4

vertices

f (x) = x^{2} - 5 x + 4

Minimum (\frac{5}{2}, - \frac{9}{4})

Solution steps

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

The parabola parameters are:

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

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

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

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{9}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{5}{2}, - \frac{9}{4})

x^{2} + 4

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

y^{2} - 2 y - 8 x - 31 = 0

x^{2} - x - 6

x = 4 y^{2}

What is the vertices f(x)=x^2-5x+4 ?
The vertices f(x)=x^2-5x+4 is Minimum (5/2 ,-9/4)