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

The vertices x^2-2x+2 is Minimum (1,1)

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vertices x^2-2x+2

vertices

x^{2} - 2 x + 2

Minimum (1, 1)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = 2

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 1)

f (x) = - 0.01 x^{2} + 0.7 x + 6

y = + x^{2} - 3

f (x) = (x - 1)^{2} - 4

x^{2} - x - 20

x^{2} - 8 x + 9