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

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

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vertices f(x)=2x^2-4x+1

vertices

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

Minimum (1, - 1)

Solution steps

y = 2 x^{2} - 4 x + 1

The parabola parameters are:

a = 2, b = - 4, c = 1

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

x_{v} = - \frac{( - 4 )}{2 \cdot 2}

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 = 2

Minimum (1, - 1)

y = - x^{2} + 10 x

f (x) = 12 x - 2 x^{2}

+ 1 + x^{2}

y = (x - 3)^{2} - 2

x^{2} - 5 x - 36

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