Kenneth S. answered • 09/08/17
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Consider two cases: y = 1 - |x|
Begin by graphing y=|x| which is V-shaped perpendicular rays beginning at origin & passing through quadrants I and II with slopes 1 and -1. This covers all cases where |y| = y. Now reflect it in the x-axis (is graph of y = -|x|. Now translate this upward 1 unit; the V-shaped intersection is at (0,1) and the rays head downward &to the left & right.
Next case, where |y| is replaced by -y; -y = 1 - |x|. This will be the opposite of the graph in the first case. Its v-shaped intersection is at (0,-1) and the rays head upward & to left and right.
Work this out yourself and graph it, and test it with various valid (x,y) points.
Kenneth S.
square. That's right, because |x| must not ever exceed 1 because if it were to do so, the right side would become negative, and |y| cannot equal a negative number. So the testing process of any x values whose absolute value > 1 reveals necessary prohibitions.
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09/08/17
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