What is the vertices f(x)=(x-4)^2-4 ?

The vertices f(x)=(x-4)^2-4 is Minimum (4,-4)

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vertices f(x)=(x-4)^2-4

vertices

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

Minimum (4, - 4)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} - 8 x + 12

The parabola parameters are:

a = 1, b = - 8, c = 12

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

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

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(4, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 4)

x^{2} + 8 x + 12

x^{2} + 8 x + 13

y = + x^{2} - 2 x - 6

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

x^{2} + 16 y = 0

What is the vertices f(x)=(x-4)^2-4 ?
The vertices f(x)=(x-4)^2-4 is Minimum (4,-4)