What is the vertices (x+2)^2=-12(y-5) ?

The vertices (x+2)^2=-12(y-5) is Maximum (-2,5)

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vertices (x+2)^2=-12(y-5)

vertices

(x + 2)^{2} = - 12 (y - 5)

Maximum (- 2, 5)

Solution steps

(x + 2)^{2} = - 12 (y - 5)

Rewrite

(x + 2)^{2} = - 12 (y - 5)

in the form

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

y = - \frac{x ^{2}}{12} - \frac{x}{3} - \frac{1}{3} + 5

The parabola parameters are:

a = - \frac{1}{12}, b = - \frac{1}{3}, c = - \frac{1}{3} + 5

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

x_{v} = - \frac{( - \frac{1}{3} )}{2 ( - \frac{1}{12} )}

Simplify

- \frac{- \frac{1}{3}}{2 ( - \frac{1}{12} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{12}

Maximum (- 2, 5)

2 x^{2} - 8 x + 6

f (x) = x^{2} + 14 x - 72

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

f (x) = x^{2} + 7

f (x) = x^{2} + 9

What is the vertices (x+2)^2=-12(y-5) ?
The vertices (x+2)^2=-12(y-5) is Maximum (-2,5)