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

The vertices 2x^2+2x+3y=0 is Maximum (-1/2 , 1/6)

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

vertices

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

Maximum (- \frac{1}{2}, \frac{1}{6})

Solution steps

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

Rewrite

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

in the form

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

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

The parabola parameters are:

a = - \frac{2}{3}, b = - \frac{2}{3}, c = 0

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

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

Simplify

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

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = \frac{1}{6}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{1}{6})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{2}{3}

Maximum (- \frac{1}{2}, \frac{1}{6})

y = 4 x - x^{2}

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

x^{2} + 5 x + 6

f (x) = 2 x^{2} + 6 x - 1

y^{2} = 12 x

What is the vertices 2x^2+2x+3y=0 ?
The vertices 2x^2+2x+3y=0 is Maximum (-1/2 , 1/6)