# Sketch the following functions and identify its domain and range. f(x) = 3 x>=2 x^2-4x+6 0<x<2Show complete solution and explain the answer.

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The domain of the function f(x)=3 is the interval [2,+oo).

The function is constant and it is a line that is parallel to x axis and the y intercept of the line is 3.

The range of this function is the set that comprises the element {3}.

The domain of the second function is the opened interval (0,2). To determine the range, we'll solve the expressions:

For x = 0 => f(0) = 6

For x = 2 => f(2) = 4 - 8 + 6 = 2

The range of the function is the opened interval (2;6).

The graph of the function is a branch of an upward concave parabola.

We'll determine the x intercepts. For this reason, we'll solve the equation x^2-4x+6 = 0. We'll complete the square.

(x^2 - 4x + 4) -4 + 6 = 0

(x-2)^2 + 2 = 0

(x-2)^2 = -2

Since a real number raised to square yields always a positive value, the quadratic equation has no real solutions. Therefore, the parabola is located above x axis, never intercepting it.

The minimum point of parabola is: V(-b/2a ; -delta/4a)

a = 1 , b = -4 , c = 6

delta = b^2-4ac = 16 - 24 = -8

V(4/2 ; 8/4) = V(2 ; 2)

The graph of the branch of parabola is:

CORRECTION:

f(x) = 3 x>=2

x^2-4x+6 0<x<2

[|x|] x<=0