What is the vertices 2x^2-5x+3 ?

The vertices 2x^2-5x+3 is Minimum (5/4 ,-1/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2x^2-5x+3

vertices

2 x^{2} - 5 x + 3

Minimum (\frac{5}{4}, - \frac{1}{8})

Solution steps

y = 2 x^{2} - 5 x + 3

The parabola parameters are:

a = 2, b = - 5, c = 3

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{1}{8}

Therefore the parabola vertex is

(\frac{5}{4}, - \frac{1}{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{1}{8})

2 X^{2} + 3 X - 20

2 X^{2}

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

y = 5 x^{2} + 210 x - 34775

x^{2} - 3 x - 10