f(x) = |x-4| + 2
|a| ≥ 0 ∀ a∈R
That means the minim of the function will be when |x-4|=0 ⇒ x = 4
f(4) = |4-4|+2
f(4) = 2 ⇒ M(4,2) - minim point
We notice that the function f is ascending for x ∈ [4,+∞) and is descending for x ∈ (-∞,4]
Because the function is linear,the sketch of the graph will be 2 lines with the intersection in the point M(4,2) (in our case).
Let's find f(0) = |0-4|+2 = 4+2 = 6 ⇒ N (0,6) is the intersection of the function with Oy axis. (now we can sketch the graph only for x ∈(-∞,4] )
Now let's calculate the function for a value bigger than 4 : for example f(5) = 3 ⇒ P (5,3) ∈ Gf
So you can see the graph in the picture.Have a nice day :)