What is the vertices (x-12)^2-156 ?

The vertices (x-12)^2-156 is Minimum (12,-156)

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vertices (x-12)^2-156

vertices

(x - 12)^{2} - 156

Minimum (12, - 156)

Solution steps

y = (x - 12)^{2} - 156

Rewrite

y = (x - 12)^{2} - 156

in the form

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

y = x^{2} - 24 x - 12

The parabola parameters are:

a = 1, b = - 24, c = - 12

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

x_{v} = - \frac{( - 24 )}{2 \cdot 1}

Simplify

x_{v} = 12

Plug in

x_{v} = 12

to find the

y_{v}

value

y_{v} = - 156

Therefore the parabola vertex is

(12, - 156)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (12, - 156)

y^{2} = - 4 x

y^{2} = 4 x

(y - 7)^{2} = - 12 (x + 1)

x^{2} = - 16 y

x^{2}

What is the vertices (x-12)^2-156 ?
The vertices (x-12)^2-156 is Minimum (12,-156)