y = |x| is two half-lines meeting at the origin, forming a V. y = a|x − h| + k moves the corner to (h, k) and sets the steepness; a negative a turns the V over.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Try it first
A machine is set to cut rods 50 cm long. A rod’s error is how far its length is from 50 cm, counted the same either way. Five rods measure 46, 48, 50, 52 and 54 cm. Plotting each error above its length, what shape do the five points make?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Two lines in disguise
|x| is x when x ≥ 0 and −x when x < 0. So the V is the line y = x on the right and y = −x on the left, meeting at the corner. Slopes 1 and −1.
Moving the vertex
y = |x − h| + k moves the corner to (h, k), the inside term running backwards as always. |x + 1| − 3 has its corner at (−1, −3).
Turning it over
A negative outside flips the V upside down, so the vertex becomes a maximum. y = −|x| + 5 peaks at (0, 5).
Evaluating
y = |x − 3| + 2 at x = 1: |−2| + 2 = 4. At x = 7: |4| + 2 = 6. The two arms give the same output at equal distances from the corner.
Piecewise generally
A piecewise function uses different rules on different stretches. Decide which condition the input satisfies before computing anything. The absolute value is the simplest piecewise function there is.
Solving with it
|x − 3| + 2 = 6 gives |x − 3| = 4, so x − 3 = 4 or x − 3 = −4: x = 7 or x = −1. Two answers, one on each arm.
Step 2: Try It Yourself
Tap and try it out.
Solve |x − 3| + 2 = 6
- 1|x − 3| + 2 = 6the equation as given
Step 0 of 4
Step 3: In Real Life
A parking garage and a thermostat
A garage charges $5 for the first hour, $3 an hour after, capped at $20: three rules on three stretches. A thermostat minimizes |T − 70|, the V-shaped distance from its target.
Step 4: Watch an Example
One step at a time.
Watch Priya Solve 2|x + 1| − 3 = 7
A fresh equation with the bars buried under two layers, peeled back one layer at a time, split into its two arms, and checked.
- Step 1
The bars are buried here: a 2 multiplies them and a 3 is taken off outside them. Nothing useful can be said about the inside until |x + 1| stands on its own, because only then is the distance from zero a known number. So the outside layers come off first, in the order that undoes them.
Watch Leo Solve 15 − 3|x − 4| = 6
The same shape of equation, with a negative multiplier this time. Leo peels the two outside layers back, and you take it from there.
- Step 1
Taking 15 from both sides leaves −3|x − 4| = −9. The bars still have a multiplier attached, so the inside is not pinned down yet.
Watch Ana Use a Three-Piece Rule
f(x) = −2x for x < 0, f(x) = 5 for 0 ≤ x < 4, and f(x) = x + 3 for x ≥ 4. Ana evaluates it at three inputs, one of them a boundary.
- Step 1
Three rules, three stretches of the number line, and no input falls under two of them. So the first job at any input is settling which stretch it lands in; the arithmetic comes after that, and only from the one rule that owns it.
Step 5: Your Turn
Practice makes it stick.
The Takings Bonus
Problem 1 of 2
A shop aims to take $400 a day: too little is lost trade, too much means it ran out of stock early. The manager’s bonus, in dollars, is b = 50 − |t − 400|, where t is the day’s takings in dollars. What is the bonus on a day when the shop takes $372?
dollars
The Data Plan
Problem 2 of 2
A phone plan costs $20 a month for anything up to 5 gigabytes of data. Above that, it costs $20 plus $4 for each whole gigabyte over 5. What does a month using 9 gigabytes cost, in dollars?
dollars
Corners, Values and Arms
The corner sits where the inside of the bars is zero. On a piecewise rule, settle the condition before the arithmetic.
1 of 5
y = |x + 6| − 4. What is the x-coordinate of the corner?
2 of 5
y = |x − 2| + 5. What is the smallest value y can take?
3 of 5
f(x) = 5 − x for x < 3, and f(x) = 2x for x ≥ 3. What is f(3)?
4 of 5
Solve 3|x − 2| = 12. Which pair of values does it give?
5 of 5
For x ≥ 4 the right arm of y = |x − 4| + 1 is a straight line. Write that line’s equation.
Whose Step?
One line on each page is where the working goes astray. Find it, then put it right.
1 of 4
Maya solved |x − 5| + 3 = 8 and finished with x = 10 as the only answer. One line is where it goes astray. Which?
Maya’s solution of |x − 5| + 3 = 8
- 1|x − 5| + 3 = 8
- 2|x − 5| = 5− 3
- 3x − 5 = 5
- 4x = 10+ 5
2 of 4
What is the other solution of |x − 5| + 3 = 8?
3 of 4
Dev was given f(x) = x + 7 for x < 2, and f(x) = 4x for x ≥ 2. Asked for f(2), Dev wrote 9. One line is where it goes astray. Which?
Dev’s working for f(2)
- 1f(x) = x + 7 for x < 2; f(x) = 4x for x ≥ 2
- 2x = 2, so use x + 7
- 3f(2) = 2 + 7
- 4f(2) = 9
4 of 4
What is f(2)?
Pick the Move
Decide what to do first, and why. No full solution is needed.
1 of 2
Ravi is about to solve 3|x − 1| + 4 = 19. Which first move sets up working that finds both answers?
2 of 2
Maya is asked how many solutions |x − 4| + 5 = 3 has. Which move settles it, and what does it give?
Same Thing?
Judge these without drawing anything: which really say the same thing?
1 of 2
Select every equation whose solutions are exactly x = 1 and x = 7.
2 of 2
Select every pair that describes the same graph.
Step 6: Quick Check
Show what you know.
Question 1 of 1
y = |x − 7| + 3. What is the minimum value?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which number k does |x − 3| + k = 7 have exactly one solution?
Stretch 2 of 2
f(x) = x + c for x < 2, and f(x) = 3x for x ≥ 2. For which number c does the graph join up at x = 2, with no jump?
What You Learned
- y = |x| is a V: the lines y = x and y = −x meeting at the origin.
- The same transformation rules move and flip it as any other function.
- To solve with an absolute value, isolate it, then take both arms.
- A piecewise function needs the right condition chosen before anything is computed.
For the grown-up
The misconception this lesson targets is that an absolute value equation has one answer. |x − 5| = 5 asks which numbers sit 5 away from zero, and two of them do, so a solution on each arm of the V is the ordinary case rather than a curiosity. The slip is unusually durable because the answer a student keeps is genuinely correct: checking it in the original equation raises no alarm, and only counting the answers catches the loss. Two cousins travel with it — dropping the bars before whatever multiplies or adds outside them has been undone, and letting a boundary such as x = 2 belong to both pieces of a piecewise rule at once. One question defeats all three, asked out loud before any arithmetic: what must the inside of the bars be, and which single condition does this input satisfy?