What is the vertices (x+4)^2=12y-24 ?

The vertices (x+4)^2=12y-24 is Minimum (-4,2)

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vertices (x+4)^2=12y-24

vertices

(x + 4)^{2} = 12 y - 24

Minimum (- 4, 2)

Solution steps

(x + 4)^{2} = 12 y - 24

Rewrite

(x + 4)^{2} = 12 y - 24

in the form

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

y = \frac{1 \cdot x ^{2}}{12} + \frac{2 x}{3} + \frac{10}{3}

The parabola parameters are:

a = \frac{1}{12}, b = \frac{2}{3}, c = \frac{10}{3}

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

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

Simplify

- \frac{\frac{2}{3}}{2 ( \frac{1}{12} )} : - 4

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 4, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{12}

Minimum (- 4, 2)

x^{2} - 10 x - 4 y + 21 = 0

f (x) = - 3 x^{2}

f (x) = - x^{2} + 4 x

f (x) = 108 - 5 x^{2} - 7 x

2 x^{2} - x - 1

What is the vertices (x+4)^2=12y-24 ?
The vertices (x+4)^2=12y-24 is Minimum (-4,2)