What is the vertices 3x^2+8y=0 ?

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

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vertices 3x^2+8y=0

vertices

3 x^{2} + 8 y = 0

Maximum (0, 0)

Solution steps

3 x^{2} + 8 y = 0

Rewrite

3 x^{2} + 8 y = 0

in the form

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

y = - \frac{3 x ^{2}}{8}

The parabola parameters are:

a = - \frac{3}{8}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - \frac{3}{8} )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{3}{8}

Maximum (0, 0)

3 x^{2} + 2 x - 3

2 x^{2} + 4 x + 1

- x^{2} + 2 x + 8

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

y = - 2 x^{2} - 4 x + 6