What is the vertices f(x)=x^2-2x+1 ?

The vertices f(x)=x^2-2x+1 is Minimum (1,0)

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vertices f(x)=x^2-2x+1

vertices

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

Minimum (1, 0)

Solution steps

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

The parabola parameters are:

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

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} = 0

Therefore the parabola vertex is

(1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 0)

y^{2} = - 10 x

f (x) = x^{2} - 6 x + 5

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x^{2} + y = 1

x = y^{2} - 3

What is the vertices f(x)=x^2-2x+1 ?
The vertices f(x)=x^2-2x+1 is Minimum (1,0)