What is the vertices V(x)=325x^2-4600x+145000 ?

The vertices V(x)=325x^2-4600x+145000 is Minimum (92/13 , 1673400/13)

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vertices V(x)=325x^2-4600x+145000

vertices

V (x) = 325 x^{2} - 4600 x + 145000

Minimum (\frac{92}{13}, \frac{1673400}{13})

Solution steps

y = 325 x^{2} - 4600 x + 145000

The parabola parameters are:

a = 325, b = - 4600, c = 145000

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

x_{v} = - \frac{( - 4600 )}{2 \cdot 325}

Simplify

- \frac{- 4600}{2 \cdot 325} : \frac{92}{13}

x_{v} = \frac{92}{13}

Plug in

x_{v} = \frac{92}{13}

to find the

y_{v}

value

y_{v} = \frac{1673400}{13}

Therefore the parabola vertex is

(\frac{92}{13}, \frac{1673400}{13})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 325

Minimum (\frac{92}{13}, \frac{1673400}{13})

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

y = 2 x^{2} - 6 x + 4

y = x^{2} - 16

y = x^{2} - 13

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

What is the vertices V(x)=325x^2-4600x+145000 ?
The vertices V(x)=325x^2-4600x+145000 is Minimum (92/13 , 1673400/13)