What is the vertices 6x^2+12y-18=0 ?

The vertices 6x^2+12y-18=0 is Maximum (0, 3/2)

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vertices 6x^2+12y-18=0

vertices

6 x^{2} + 12 y - 18 = 0

Maximum (0, \frac{3}{2})

Solution steps

6 x^{2} + 12 y - 18 = 0

Rewrite

6 x^{2} + 12 y - 18 = 0

in the form

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

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

The parabola parameters are:

a = - \frac{1}{2}, b = 0, c = \frac{3}{2}

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

x_{v} = - \frac{0}{2 ( - \frac{1}{2} )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = \frac{3}{2}

Therefore the parabola vertex is

(0, \frac{3}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{2}

Maximum (0, \frac{3}{2})

x^{2} + 8 y + 2 x - 23 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

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

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

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

What is the vertices 6x^2+12y-18=0 ?
The vertices 6x^2+12y-18=0 is Maximum (0, 3/2)