What is the vertices (x+2)^2=-8(y+3) ?

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

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vertices (x+2)^2=-8(y+3)

vertices

Maximum (- 2, - 3)

Solution steps

(x + 2)^{2} = - 8 (y + 3)

Rewrite

(x + 2)^{2} = - 8 (y + 3)

in the form

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

y = - \frac{x ^{2}}{8} - \frac{x}{2} - \frac{1}{2} - 3

The parabola parameters are:

a = - \frac{1}{8}, b = - \frac{1}{2}, c = - \frac{1}{2} - 3

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

x_{v} = - \frac{( - \frac{1}{2} )}{2 ( - \frac{1}{8} )}

Simplify

- \frac{- \frac{1}{2}}{2 ( - \frac{1}{8} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 2, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{8}

Maximum (- 2, - 3)

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