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

The vertices f(x)=x^2-4x-4 is Minimum (2,-8)

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

vertices

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

Minimum (2, - 8)

Solution steps

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

The parabola parameters are:

a = 1, b = - 4, c = - 4

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(2, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 8)

f (x) = 2 x^{2} + 6 x + 1

x^{2} + 10 x + 11

y = x^{2} + 2 x - 35

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

f (x) = x^{2} - 10 x + 9

What is the vertices f(x)=x^2-4x-4 ?
The vertices f(x)=x^2-4x-4 is Minimum (2,-8)