What is the vertices-(x^2)/(10)+(9x)/(10)+11/5 ?

The vertices-(x^2)/(10)+(9x)/(10)+11/5 is Maximum (9/2 , 169/40)

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vertices-(x^2)/(10)+(9x)/(10)+11/5

vertices

- \frac{x ^{2}}{10} + \frac{9 x}{10} + \frac{11}{5}

Maximum (\frac{9}{2}, \frac{169}{40})

Solution steps

y = - \frac{x ^{2}}{10} + \frac{9 x}{10} + \frac{11}{5}

The parabola parameters are:

a = - \frac{1}{10}, b = \frac{9}{10}, c = \frac{11}{5}

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

x_{v} = - \frac{( \frac{9}{10} )}{2 ( - \frac{1}{10} )}

Simplify

- \frac{\frac{9}{10}}{2 ( - \frac{1}{10} )} : \frac{9}{2}

x_{v} = \frac{9}{2}

Plug in

x_{v} = \frac{9}{2}

to find the

y_{v}

value

y_{v} = \frac{169}{40}

Therefore the parabola vertex is

(\frac{9}{2}, \frac{169}{40})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{10}

Maximum (\frac{9}{2}, \frac{169}{40})

x^{2} + y^{2} = 17

y^{2} = x + 1

4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

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

9 x^{2} + 16 y^{2} = 144

What is the vertices-(x^2)/(10)+(9x)/(10)+11/5 ?
The vertices-(x^2)/(10)+(9x)/(10)+11/5 is Maximum (9/2 , 169/40)