Algebra 1
Absolute value: graphs & distance
Sometimes what you care about is how far, not which direction. A part machined to "within 0.5 millimeters of target" can be a little over or a little under. What matters is the size of the gap, not its sign. That idea, distance with the direction thrown away, is exactly what absolute value measures.
Picture the number line. The absolute value of a number, written |x|, is simply how far x sits from 0, and distance is never negative. So |5| = 5, because 5 is five steps from 0. And |−5| = 5 too, because −5 is also five steps from 0, just on the other side. Different inputs, same distance. (This builds on the number line from Unit 1; here it becomes a tool.)
Once you read |x| as distance, its graph almost draws itself. Plot y = |x| at a few inputs: at x = −3 you get 3, at −2 you get 2, at −1 you get 1, at 0 you get 0, then 1, 2, 3 going right. Those points are (-3,3), (-2,2), (-1,1), (0,0), (1,1), (2,2), (3,3), and joined up they make a V sitting on the origin 8.3.f1.
The right arm climbs at slope +1, the left arm climbs at slope −1, and the whole thing rests on the x-axis, never dipping below it, because an output that is a distance can't be negative. Shifts just slide that V around without changing its shape: y = |x| + 2 lifts it up 2, y = |x| − 3 drops it down 3, and y = |x − 1| slides it right 1, so its vertex moves to where the inside equals 0. Those are pictures to read off, not equations to solve.
Now the solving, read straight from the distance picture. There are three shapes, and each one is a plain-language question about distance from 0:
- |x| = k asks "which numbers are exactly k from 0?" Two of them, one each way: x = k or x = -k. (This needs k ≥ 0. At k = 0 the two answers collapse to the single x = 0; if k is negative there's no solution, since no distance is negative.)
- |x| < k asks "which numbers are within k of 0?" Everything in the band around 0: -k < x < k. That's an and, a single segment, and it's a direct payoff of Lesson 8.2. (This needs k > 0.)
- |x| > k asks "which numbers are more than k from 0?" Everything out past k on either side: x < -k or x > k. That's an or, two rays. (This holds for any k ≥ 0.)
One setup step comes before you read any of those off: get the absolute value by itself first. If it's scaled, as in 2|x| ≤ 8, divide to isolate it (here divide by 2 to get |x| ≤ 4), and then read the distance. Keep that isolating step a division by a positive number for now.
New words
- 8.3.d1 Absolute value |x|: the distance of x from 0 on the number line; always ≥ 0. |5| = 5 and |-5| = 5, both are 5 units from 0 (callback to Unit 1.4).
- 8.3.d2 The graph y = |x|: a V with its vertex at the origin (the right arm rises at slope +1, the left arm at slope -1), and it never dips below the x-axis, because an output that is a distance can't be negative.
- 8.3.d3 Simple shifts of the V: y = |x| + 2 lifts the whole V up 2; y = |x| − 3 drops it 3; y = |x − 1| slides it right 1 (the vertex moves to where the inside equals 0). These are graph moves read off the picture, not algebra to solve.
Worked example
8.3.w1 Example: inclusive "within". Solve |x| ≤ 2. This asks for the numbers within 2 of 0, and the ≤ includes the boundary, so -2 ≤ x ≤ 2. Filled circles at -2 and 2, shade the segment between. Solution: -2 ≤ x ≤ 2.
8.3.w2 Example: the graph y = |x|. Plot (-2,2), (-1,1), (0,0), (1,1), (2,2): a V with vertex (0,0), arms at slope ±1, resting on the x-axis. Then y = |x| − 2 is the same V dropped 2, with vertex (0,-2), and y = |x − 1| is it slid right 1, with vertex (1,0). Nothing to solve here, you're just reading the picture.
8.3.w3 Example: "outside" (or). Solve |x| > 2. This asks for the numbers more than 2 from 0, which is x < -2 or x > 2. Open circles at -2 and 2, shade outward in two rays.
8.3.w4 Example: isolate first. Solve 2|x| ≤ 8. Divide both sides by 2 first; it's a positive 2, so nothing flips: |x| ≤ 4. That's the numbers within 4 of 0, so -4 ≤ x ≤ 4. Filled circles at -4 and 4, shade between.
8.3.w5 Example: no solution / all reals. Distance is never negative, so you can read these straight off the picture without any algebra. |x| = -3 has no solution, because nothing sits a negative distance from 0. |x| < -2 has no solution for the same reason: no distance is below -2. And |x| > -1 is true for every real x, since every distance is more than -1.
That last example points at the trap to watch here. It's tempting to give |x| = k a single answer and stop, but distance runs both ways, so expect two: x = k and x = −k (the lone exception is k = 0).
And a quick self-check on the and/or split: "within" or "less than" gives one segment between -k and k, while "outside" or "greater than" gives two rays. If you're ever unsure which, don't reach for a memorized rule. Go back to the distance question and re-derive it.
Check yourself
- 8.3.c1 Sketch y = |x| in words: where's the vertex, what are the arm slopes, and why does it never go below the x-axis? Then describe y = |x| + 1. (Vertex at the origin (0,0); right arm slope +1, left arm slope −1; it stays at or above the axis because an output that is a distance can't be negative. y = |x| + 1 is the same V lifted up 1, vertex at (0,1).)
- 8.3.c2 Why does |x| < 4 become an and (a single segment) while |x| > 4 becomes an or (two rays)? Use the distance picture. ("Within 4 of 0" means closer than 4 on both sides at once: an and, the band -4 < x < 4. "More than 4 from 0" means out past 4 on either side: an or, the two rays x < -4 or x > 4.)
- 8.3.c3 What is the solution of |x| = -3? Of |x| > -3? Explain each from "distance is never negative." (|x| = -3 has no solution: nothing sits a negative distance from 0. |x| > -3 is all real numbers: every distance is greater than -3, so every x qualifies.)
For practice, stay on the core path: everything here is centered at 0, so you can read each one off the distance picture. Give both solutions for the equations, describe the graph for the inequalities, and check your answers by substituting.
Practice problems (core, the mastery path): symmetric, centered at 0. (Problems 2, 3, 5, 6, 7 shift the center off 0, so they've moved to the Reach set below.)
Equations (give both solutions; check):
Inequalities (solve; describe the graph):
AnswersTry each one yourself first, then open to check.
- x = 4 or x = -4. (|4| = 4, |-4| = 4.)
- x = 4 or x = -4. (|2·4| = 8, |2·(-4)| = 8.)
- x = 7 or x = -7. (|7| = 7, |-7| = 7.)
- -3 < x < 3 — |2x| < 6 ⇒ -6 < 2x < 6; open circles at -3 and 3, shade between (and).
- x < -5 or x > 5 — "more than 5 from 0"; open circles at -5 and 5, shade outward (or).
- -5 ≤ x ≤ 5 — isolate: divide by 2 → |x| ≤ 5; filled circles at -5 and 5, shade between.
- no solution — a distance can't be less than -2.
- all real numbers — every distance is more than -1.
Reach beyond Algebra 1 (optional, off the mastery path)
This part is here so you can see where the idea goes next. It isn't part of the Algebra 1 target, so get the distance-from-0 core above solid first and treat the rest as a preview.
When the center moves off 0, as in |x − 3| = 5 or |2x − 1| = 7, you can't just read x = ±k anymore. The move is to split into two cases (the inside equals +k, or the inside equals −k), solve each, and name the answer set. Here's that method on one example.
8.3.w6 w1: |x − 3| = 5. Two cases: x − 3 = 5 → x = 8, or x − 3 = -5 → x = -2. Solutions: x = 8 or x = -2. Check: |8 − 3| = 5, |-2 − 3| = 5. (Pictured: "distance 5 from 3.")
A few more, same idea. 8.3.w7 w2: |x − 3| < 5. Within 5 of 3 → -5 < x − 3 < 5 → -2 < x < 8 (interval (-2, 8)). 8.3.w8 w3: |x − 3| > 5. More than 5 from 3 → x − 3 < -5 or x − 3 > 5 → x < -2 or x > 8. 8.3.w9 w4: |x + 2| = 3. |x + 2| = |x − (-2)|, distance 3 from -2 → x = 1 or x = -5.
Reach practice (optional):
- |x + 2| = 3
- |x - 1| = 4
- |2x - 1| = 7
- |x + 1| < 3
- |x - 2| ≥ 4
Reach answer key:
- x = 1 or x = -5. (|1+2| = 3, |-5+2| = 3.)
- x = 5 or x = -3. (|5-1| = 4, |-3-1| = 4.)
- x = 4 or x = -3. (2x-1 = 7 → x = 4; 2x-1 = -7 → x = -3.)
- -4 < x < 2 — "within 3 of -1"; interval (-4, 2).
- x ≤ -2 or x ≥ 6 — "at least 4 from 2"; two rays.