What is the vertices y=x^2-2x+3 ?

The vertices y=x^2-2x+3 is Minimum (1,2)

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vertices y=x^2-2x+3

vertices

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

Minimum (1, 2)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = 3

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} = 2

Therefore the parabola vertex is

(1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 2)

y = + (2 x + 1)^{2} - 1

(x + 2)^{2} + 3

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

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

- x^{2} - x + 2