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

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

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

vertices

f (x) = - \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})

y^{2} = 32 x

y + 4 x \leq x^{2} + 2

y^{2} - y - x + 1 = 0

9 y^{2} - 16 x^{2} = 81

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

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