What is the vertices y=x^2-2x+5 ?

The vertices y=x^2-2x+5 is Minimum (1,4)

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vertices y=x^2-2x+5

vertices

y = x^{2} - 2 x + 5

Minimum (1, 4)

Solution steps

y = x^{2} - 2 x + 5

The parabola parameters are:

a = 1, b = - 2, c = 5

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} = 4

Therefore the parabola vertex is

(1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 4)

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

x^{2} - y = 0

y = x^{2} + 8 x + 12

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

f (x) = - (x + 3) (x + 10)