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

The vertices 3x^2+3x+2y=0 is Maximum (-1/2 , 3/8)

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

vertices

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

Maximum (- \frac{1}{2}, \frac{3}{8})

Solution steps

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

Rewrite

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

in the form

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

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

The parabola parameters are:

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

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

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

Simplify

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

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{3}{8}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{3}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{3}{2}

Maximum (- \frac{1}{2}, \frac{3}{8})

f (x) = 2 x^{2} + 8 x + 3

y = 2 x^{2} - 7

y = - 3 x^{2} - 36 x - 104

f (x) = x^{2} - 6 x + 11

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

What is the vertices 3x^2+3x+2y=0 ?
The vertices 3x^2+3x+2y=0 is Maximum (-1/2 , 3/8)