What is the vertices f(x)=x^2-2x+5 ?

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

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vertices f(x)=x^2-2x+5

vertices

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

f (x) = x^{2} - 3 x + 10

- 2 x^{2} + 12 x - 20

y^{2} + 2 x + 6 y + 17 = 0

f (x) = 2 x^{2} + 3 x + 1

What is the vertices f(x)=x^2-2x+5 ?
The vertices f(x)=x^2-2x+5 is Minimum (1,4)