What is the vertices f(x)=x^2-24x-12 ?

The vertices f(x)=x^2-24x-12 is Minimum (12,-156)

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vertices f(x)=x^2-24x-12

vertices

f (x) = x^{2} - 24 x - 12

Minimum (12, - 156)

Solution steps

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)

x^{2} = - 12 y

f (x) = x^{2} + 6 x + 8

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

x = (y + 2)^{2}

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

What is the vertices f(x)=x^2-24x-12 ?
The vertices f(x)=x^2-24x-12 is Minimum (12,-156)