Modulus • Topic 3 of 3

Modulus Graphs

The graph of y = |x| is a V with its vertex at the origin, made of the line y = x for x ≥ 0 and y = −x for x < 0. Every modulus graph is built from this V by shifting and stretching. For y = |x − a| + b the vertex moves to (a, b): subtracting a inside slides the V right by a, adding b outside lifts it up by b. The two arms always have slopes +1 and −1 (a coefficient like |2x − 4| = 2|x − 2| steepens them to ±2). Reading the graph is the fastest way to answer "how many solutions" questions: the number of times a horizontal line y = c meets the V tells you how many roots |x − a| + b = c has — two if c > b, one if c = b, none if c < b. The minimum value of |x − a| + b is b, reached at x = a, which is exactly why these graphs are gold for maxima–minima questions in CAT.

✅ Solved examples

1. Where is the vertex of y = |x − 3| + 2, and what is its minimum value?
The form |x − a| + b has vertex (a, b) = (3, 2). Minimum value is 2, attained at x = 3.
2. For y = |x + 1| − 4, find the x-intercepts.
Set y = 0: |x + 1| = 4 ⇒ x + 1 = ±4 ⇒ x = 3 or x = −5. The V crosses the x-axis at (3, 0) and (−5, 0).
3. How many solutions does |x − 2| + 1 = 4 have? Use the graph.
Vertex of y = |x − 2| + 1 is (2, 1); the line y = 4 sits above the vertex (4 > 1), so it cuts both arms ⇒ 2 solutions. (Indeed |x − 2| = 3 ⇒ x = 5 or −1.)
4. Find the minimum value of y = |x − 1| + |x − 5|.
This is the total distance from x to 1 and to 5, minimised for any x between 1 and 5, giving the gap |1 − 5| = 4. The graph is flat at height 4 across [1, 5]. Minimum = 4.

✏️ Practice — try these, take hints as needed

1. State the vertex of y = |x + 2| − 3.
Write as |x − a| + b.
x + 2 = x − (−2), so a = −2.
b = −3.
(−2, −3)
2. Find the minimum value of y = |x − 7| + 5.
Minimum of |x − a| + b is b.
Here b = 5.
Reached at x = 7.
5 (at x = 7)
3. How many roots does |x| = 3 − x have? (Think graphically.)
Graph y = |x| and line y = 3 − x.
For x ≥ 0: x = 3 − x.
For x < 0: −x = 3 − x has no x.
1 (x = 1.5)
4. Find the x-intercepts of y = |2x − 6|.
Set the modulus to 0.
2x − 6 = 0.
Single touch point.
x = 3 (the vertex touches the axis)
5. For what values of c does |x − 4| + 2 = c have exactly two solutions?
Vertex height is 2.
Two cuts need c above the vertex.
c > 2.
c > 2

📝 Topic test — 8 questions

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