What is the vertices 6x^2-13x+6 ?

The vertices 6x^2-13x+6 is Minimum (13/12 ,-25/24)

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vertices 6x^2-13x+6

vertices

6 x^{2} - 13 x + 6

Minimum (\frac{13}{12}, - \frac{25}{24})

Solution steps

y = 6 x^{2} - 13 x + 6

The parabola parameters are:

a = 6, b = - 13, c = 6

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

x_{v} = - \frac{( - 13 )}{2 \cdot 6}

Simplify

x_{v} = \frac{13}{12}

Plug in

x_{v} = \frac{13}{12}

to find the

y_{v}

value

y_{v} = - \frac{25}{24}

Therefore the parabola vertex is

(\frac{13}{12}, - \frac{25}{24})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (\frac{13}{12}, - \frac{25}{24})

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

y = x^{2} - 10 x - 4

x^{2} + 6 y = 0

f (x) = 7 x^{2} - x + 1

f (x) = 3 x^{2} - 6 x + 2

What is the vertices 6x^2-13x+6 ?
The vertices 6x^2-13x+6 is Minimum (13/12 ,-25/24)