For y = 2|x − 3|, there will be no negative values of y because of the Absolute Value Bars around (x − 3).
For (x − 3) > 0, one will have the equation of a line with positive slope since both 2 and (x − 3) will be ever
positive for x > 3 in y = 2|x − 3|. The first linear equation will then amount to y = 2(x − 3).
For (x − 3) < 0, one will have the equation of a line with negative slope since (x − 3) in y = 2|x − 3|will be negative for x < 3 and y = 2|x − 3| will amount to y = -2(x − 3).
For (x − 3) = 0, y = 2(x − 3) & y = -2(x − 3) intersect at (x,y) = (3,0).
The two linear equations sought are then y = 2(x − 3) & y = -2(x − 3). An online graphing application will show that their graphs intersect at (x,y) = (3,0).
