What is the vertices x^2-6x+5y=34 ?

The vertices x^2-6x+5y=34 is Maximum (3, 43/5)

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vertices x^2-6x+5y=34

vertices

x^{2} - 6 x + 5 y = 34

Maximum (3, \frac{43}{5})

Solution steps

x^{2} - 6 x + 5 y = 34

Rewrite

x^{2} - 6 x + 5 y = 34

in the form

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

y = - \frac{x ^{2}}{5} + \frac{6 x}{5} + \frac{34}{5}

The parabola parameters are:

a = - \frac{1}{5}, b = \frac{6}{5}, c = \frac{34}{5}

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

x_{v} = - \frac{( \frac{6}{5} )}{2 ( - \frac{1}{5} )}

Simplify

- \frac{\frac{6}{5}}{2 ( - \frac{1}{5} )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = \frac{43}{5}

Therefore the parabola vertex is

(3, \frac{43}{5})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{5}

Maximum (3, \frac{43}{5})

f (x) = 5 x^{2} + 1

\frac{5 x ^{2}}{2}

x^{2} - 2 x

- x^{2} - 2 x + 3

y^{2} = 6 x - 32

What is the vertices x^2-6x+5y=34 ?
The vertices x^2-6x+5y=34 is Maximum (3, 43/5)