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

The vertices f(x)=x^2-4x+4 is Minimum (2,0)

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

vertices

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

Minimum (2, 0)

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} = 0

Therefore the parabola vertex is

(2, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 0)

y^{2} = 4 x

y = x^{2} - 2 x - 3

y = x^{2} + 2 x + 1

10 + 3 x^{2}

x^{2} - 5 x - 24

What is the vertices f(x)=x^2-4x+4 ?
The vertices f(x)=x^2-4x+4 is Minimum (2,0)