What is the vertices f(x)=415x^2-4600x+200000 ?

The vertices f(x)=415x^2-4600x+200000 is Minimum (460/83 , 15542000/83)

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vertices f(x)=415x^2-4600x+200000

vertices

f (x) = 415 x^{2} - 4600 x + 200000

Minimum (\frac{460}{83}, \frac{15542000}{83})

Solution steps

y = 415 x^{2} - 4600 x + 200000

The parabola parameters are:

a = 415, b = - 4600, c = 200000

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

x_{v} = - \frac{( - 4600 )}{2 \cdot 415}

Simplify

- \frac{- 4600}{2 \cdot 415} : \frac{460}{83}

x_{v} = \frac{460}{83}

Plug in

x_{v} = \frac{460}{83}

to find the

y_{v}

value

y_{v} = \frac{15542000}{83}

Therefore the parabola vertex is

(\frac{460}{83}, \frac{15542000}{83})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 415

Minimum (\frac{460}{83}, \frac{15542000}{83})

y = x^{2} - 4 x 3

f (x) = x^{2} - 3 x - 10

- x^{2} + 2 x - 10

f (x) = 3 x^{2} - 12 x + 13

y = x^{2} - x - 6

What is the vertices f(x)=415x^2-4600x+200000 ?
The vertices f(x)=415x^2-4600x+200000 is Minimum (460/83 , 15542000/83)