What is the vertices x^2-4x+4 ?

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

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vertices x^2-4x+4

vertices

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)

t^{2}

f (x) = - x^{2} - 6 x + 3

x^{2} - 2 x - 3

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

(x + 5)^{2}