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

The vertices+3x^2+x+5 is Minimum (-1/6 , 59/12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices+3x^2+x+5

vertices

+ 3 x^{2} + x + 5

Minimum (- \frac{1}{6}, \frac{59}{12})

Solution steps

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

The parabola parameters are:

a = 3, b = 1, c = 5

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

x_{v} = - \frac{1}{2 \cdot 3}

Simplify

x_{v} = - \frac{1}{6}

Plug in

x_{v} = - \frac{1}{6}

to find the

y_{v}

value

y_{v} = \frac{59}{12}

Therefore the parabola vertex is

(- \frac{1}{6}, \frac{59}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{1}{6}, \frac{59}{12})

y = 3 x^{2} + x - 1

28 x^{2} + 47 x + 15

y^{2} - 8 y + 4 x = 0

4 y^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

x^{2} + 3 x + 2