What is the vertices (x+2)^2=-6(y-1) ?

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

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vertices (x+2)^2=-6(y-1)

vertices

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

Maximum (- 2, 1)

Solution steps

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

Rewrite

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

in the form

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

y = - \frac{x ^{2}}{6} - \frac{2 x}{3} - \frac{2}{3} + 1

The parabola parameters are:

a = - \frac{1}{6}, b = - \frac{2}{3}, c = - \frac{2}{3} + 1

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

x_{v} = - \frac{( - \frac{2}{3} )}{2 ( - \frac{1}{6} )}

Simplify

- \frac{- \frac{2}{3}}{2 ( - \frac{1}{6} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{6}

Maximum (- 2, 1)

y = 2 x^{2} - x

- x^{2} + 14 x - 45

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

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

y = x^{2} + 16 x + 71

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