What is the vertices y=(95x^2-10x+26)/(10) ?

The vertices y=(95x^2-10x+26)/(10) is Minimum (1/19 , 489/190)

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vertices y=(95x^2-10x+26)/(10)

vertices

y = \frac{95 x ^{2} - 10 x + 26}{10}

Minimum (\frac{1}{19}, \frac{489}{190})

Solution steps

y = \frac{95 x ^{2} - 10 x + 26}{10}

Rewrite

y = \frac{95 x ^{2} - 10 x + 26}{10}

in the form

y = a x^{2} + b x + c

y = \frac{19 x ^{2}}{2} - x + \frac{13}{5}

The parabola parameters are:

a = \frac{19}{2}, b = - 1, c = \frac{13}{5}

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

x_{v} = - \frac{( - 1 )}{2 ( \frac{19}{2} )}

Simplify

- \frac{- 1}{2 ( \frac{19}{2} )} : \frac{1}{19}

x_{v} = \frac{1}{19}

Plug in

x_{v} = \frac{1}{19}

to find the

y_{v}

value

y_{v} = \frac{489}{190}

Therefore the parabola vertex is

(\frac{1}{19}, \frac{489}{190})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{19}{2}

Minimum (\frac{1}{19}, \frac{489}{190})

(x - 1)^{2} = - 2 y - 11

f (x) = x^{2} - 6 x + 10

- x^{2} + 4 x

f (x) = - 6 x^{2} + 17 x - 12

f (x) = 4 x^{2} - 16 x + 10

What is the vertices y=(95x^2-10x+26)/(10) ?
The vertices y=(95x^2-10x+26)/(10) is Minimum (1/19 , 489/190)