What is the vertices y=6x^2-29x-5 ?

The vertices y=6x^2-29x-5 is Minimum (29/12 ,-961/24)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=6x^2-29x-5

vertices

y = 6 x^{2} - 29 x - 5

Minimum (\frac{29}{12}, - \frac{961}{24})

Solution steps

y = 6 x^{2} - 29 x - 5

The parabola parameters are:

a = 6, b = - 29, c = - 5

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

x_{v} = - \frac{( - 29 )}{2 \cdot 6}

Simplify

x_{v} = \frac{29}{12}

Plug in

x_{v} = \frac{29}{12}

to find the

y_{v}

value

y_{v} = - \frac{961}{24}

Therefore the parabola vertex is

(\frac{29}{12}, - \frac{961}{24})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (\frac{29}{12}, - \frac{961}{24})

f (x) = 2 x^{2} + 7 x + 3

x = - y^{2}

f (x) = - x^{2} - 8 x - 7

4 x^{2} - 3 x + 5

f (x) = 5 x + 3 x^{2} - 4

What is the vertices y=6x^2-29x-5 ?
The vertices y=6x^2-29x-5 is Minimum (29/12 ,-961/24)