What is the vertices y=2x^2-5x+5 ?

The vertices y=2x^2-5x+5 is Minimum (5/4 , 15/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices y=2x^2-5x+5

vertices

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

Minimum (\frac{5}{4}, \frac{15}{8})

Solution steps

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

The parabola parameters are:

a = 2, b = - 5, c = 5

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

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

Simplify

x_{v} = \frac{5}{4}

Plug in

x_{v} = \frac{5}{4}

to find the

y_{v}

value

y_{v} = \frac{15}{8}

Therefore the parabola vertex is

(\frac{5}{4}, \frac{15}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{5}{4}, \frac{15}{8})

What is the vertices y=2x^2-5x+5 ?
The vertices y=2x^2-5x+5 is Minimum (5/4 , 15/8)