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

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

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

vertices

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

Minimum (2, - 2)

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(2, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 2)

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

y = 3 x^{2} + 5 x - 4

y = x^{2} + 4

x^{2} - 6 x + 5

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

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