Transforming the absolute value graph

When you reach for it

The absolute value graph is a V with its corner at the origin and arms of slope -1 and 1. The unit’s general moves, shifting, flipping and stretching, put into one rule by Combining transformations and the general form, work on it exactly as on any parent. Applied to |x| they give the family

Reach for this lesson when such a rule has to be drawn, when a V on a grid needs its rule, or when an absolute value equation or inequality is easier to see than to solve. A V is also the simplest model for a quantity that rises at a steady rate up to one point and falls at the same rate after it, such as a made-up signal meter that reads higher as you drive toward a transmitter and lower after you pass it. Real radio signals do not fade in straight lines; the V is a simplified model.

The procedure

The vertex is (h, k)

The bars |x - h| are 0 at x = h and positive at every other input. So at x = h the output is a \cdot 0 + k = k, and the corner, the vertex, sits at (h, k). The sign inside the bars works as for every parent in Graphing by transforming a parent graph: |x + 2| is |x - (-2)|, so h = -2.

The arms have slopes a and -a

To the right of the vertex x - h \ge 0, so |x - h| = x - h and the rule is the line y = a(x - h) + k, with slope a. To the left x - h < 0, so |x - h| = -(x - h) and the rule is y = -a(x - h) + k, with slope -a. Each arm is a ray. One step to either side of the vertex the height changes by a, so (h - 1, k + a) and (h + 1, k + a) are on the graph, and those three points are enough to draw it.

• a > 0: the V opens upward.
• a < 0: the V is flipped and opens downward.
• |a| > 1: narrower than |x|. |a| < 1: wider.

Domain and range

Any real number can go inside the bars, so the domain is all real numbers. Every output other than the vertex’s differs from k by a times a positive number, so:

The vertex is the graph’s only turning point, and it is the minimum or maximum of the whole function.

A factor inside the bars

A rule such as y = |2x + 6| has a factor inside. Since |mn| = |m|\,|n|, the factor comes out: |2x + 6| = |2(x + 3)| = 2|x + 3|. So for |x| a horizontal compression draws the same graph as a vertical stretch, and every V can be put in the form a|x - h| + k before anything is read.

Reading the rule from a V

The vertex gives h and k directly, as in finding the rule of a transformed graph. Any other point (x, y) on the V then gives a, since y - k = a|x - h|:

Equations and inequalities off the graph

The solutions of a|x - h| + k = c are the inputs where the horizontal line y = c meets the V, which is solving with graphs when one side is a constant. The line crosses both arms (two solutions), touches only the vertex when c = k (one solution), or misses the V (none). Two crossings sit at the same distance from h, one on each side, because the arms are mirror images in the line x = h; Symmetry about the vertical axis and the origin studies that kind of fold.

An inequality such as a|x - h| + k \le c asks where the V is on or below the line. For an upward V that is the stretch between the crossings; for a downward V it is the two pieces outside them. Absolute value equations and Absolute value inequalities reach the same answers by algebra, which is the check whenever a crossing does not land on a grid point.

Worked examples

Boundaries

• a \ne 0. With a = 0 the rule is y = k, a horizontal line with no vertex.
• (h, k) is read only from the form a|x - h| + k, with x alone inside the bars. In |2x + 6| the corner is at x = -3, not x = -6, so take the factor out first.
• The arms are straight because what sits inside the bars is linear. A rule such as |x^2 - 4| is not a V.
• A situation can limit the domain, as on Aino’s road, and the range then shrinks to the heights reached there, which need not be balanced about h.
• A crossing read off a graph is exact only when it lands on a grid point; otherwise confirm it by algebra.

Where this goes next

The V is the plainest piecewise function: two linear rules, one on each side of x = h, the pattern behind graphing any piecewise rule. Reading a vertex as (h, k) carries straight over to the vertex of a parabola in vertex form, a(x - h)^2 + k, where the sign of a again decides minimum or maximum. Meeting a V with a slanted line instead of a level one, as in Exercise 17, is the general picture of solving an equation with graphs.

Exercises

Reading the rule

For Exercises 1 to 6, give the vertex, the slopes of the left and right arms, and the range.

1. y = 2|x + 6| - 3
2. y = -|x - 2| + 8
3. \displaystyle y = -\frac{1}{4}|x| + 3
4. y = 1.5|x - 8| - 6
5. y = |4x - 12| + 2
6. y = 9 - |3x + 3|

Graphing

7. Graph y = -2|x - 1| + 4. Mark the vertex, the points one and two steps either side of it, and the intercepts.
8. Graph \displaystyle y = \frac{1}{3}|x + 3| - 2, marking the vertex and the intercepts.
9. Describe, in order, the moves that turn y = |x| into \displaystyle y = -\frac{1}{2}|x - 6| + 1. Give the vertex and the range.

Writing the rule

10. A V has its vertex at (3, -4) and passes through (5, 2). Write its rule.
11. A V has its vertex at (-4, 1) and passes through (0, -1). Write its rule.
12. A V has its vertex at (0, 6) and crosses the x-axis at x = -2 and x = 2. Write its rule.
13. A V passes through (-1, 7), (0, 5), (1, 3), (2, 1), (3, 3), (4, 5) and (5, 7). Write its rule.
14. A V has its vertex on the x-axis at x = -7, its range is y \le 0, and its arms have slopes \displaystyle \frac{2}{3} and \displaystyle -\frac{2}{3}. Write its rule.

Equations and inequalities from the graph

15. Use your graph from Exercise 7 to solve -2|x - 1| + 4 = 0, -2|x - 1| + 4 = 4, -2|x - 1| + 4 = 6 and -2|x - 1| + 4 > 0.
16. Use your graph from Exercise 8 to solve \displaystyle \frac{1}{3}|x + 3| - 2 \le 1 and \displaystyle \frac{1}{3}|x + 3| - 2 > -1.
17. Sketch y = |x + 1| and y = 2x - 4 on one grid. Solve |x + 1| = 2x - 4 and |x + 1| < 2x - 4.

Signal along a road

18. Maksim’s radio meter reads S = 24 - 2|x - 5| along a road, for 0 \le x \le 15 km. Where is the signal strongest, and how strong is it? Where is S \ge 16? What readings occur on the whole road?
19. On another road, Ilkka’s meter reads 12 at the 2 km mark and again at the 10 km mark, and it is strongest, at 20, somewhere between. Assuming the readings follow a V, write a rule for S.

Reasoning

20. Explain why the vertex of y = a|x - h| + k is the lowest point of the graph when a > 0 and the highest when a < 0.
21. Fotini says y = |-2x + 8| and y = 2|x - 4| draw the same graph. Is she right? Explain.
22. Explain how to tell from a, k and c, without solving, whether a|x - h| + k = c has two, one or no solutions. Then decide for \displaystyle \frac{1}{2}|x - 1| + 3 = 2.