What is the vertices x^2+6x+3y+2=0 ?

The vertices x^2+6x+3y+2=0 is Maximum (-3, 7/3)

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vertices x^2+6x+3y+2=0

vertices

x^{2} + 6 x + 3 y + 2 = 0

Maximum (- 3, \frac{7}{3})

Solution steps

x^{2} + 6 x + 3 y + 2 = 0

Rewrite

x^{2} + 6 x + 3 y + 2 = 0

in the form

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

y = - \frac{x ^{2}}{3} - 2 x - \frac{2}{3}

The parabola parameters are:

a = - \frac{1}{3}, b = - 2, c = - \frac{2}{3}

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

x_{v} = - \frac{( - 2 )}{2 ( - \frac{1}{3} )}

Simplify

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

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = \frac{7}{3}

Therefore the parabola vertex is

(- 3, \frac{7}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{3}

Maximum (- 3, \frac{7}{3})

y = - x^{2} + 8 x - 7

- 3 x^{2} + 6 x + 5

f (x) = - x^{2} + 12 x - 35

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

- 4 x^{2} + 6 x - 1

What is the vertices x^2+6x+3y+2=0 ?
The vertices x^2+6x+3y+2=0 is Maximum (-3, 7/3)