Today is all about the modulus function. Let's start from a line you can already draw perfectly — and then bend it.
Quick Review — activate what you already know
Four fast checks. Each one judges your answer instantly — one attempt each.
Explore — fold a line into a V
You can sketch $y = x - 2$ in seconds. Now suppose we forbid negative outputs. Watch how the graph must change.
The V you just built is the modulus function $y = |x - 2|$: keep the line where it stays above the axis, and reflect the part below it. Today we'll master this fold for any line $y = |ax + b|$, and use it to solve modulus equations like $|a| = |b|$ and inequalities like $|x - a| < b$.
Both sides have modulus $\Rightarrow$ square both sides to remove the modulus.
One side has modulus $\Rightarrow$ "distance from $a$ is less than $b$" means $x$ is between $a - b$ and $a + b$.
Interactive: adjust $a$ and $b$ to see how $y = |ax + b|$ changes.
Dashed line: $y = ax + b$ (original). Solid line: $y = |ax + b|$ (reflected V-shape).
Click a function card (left), then click the graph you think matches it. Correct pairs lock in green; wrong pairs shake red. Race to match all 6 first!
Solve the equation $|2x + 1| = |x + 5|$.
Sketch $y = |2x + 1|$ and $y = |x + 5|$ on the same axes. Where is the vertex of each V-shape? From your sketch, estimate the $x$-coordinates of the intersection points.
Currently plotting: (green) and (blue)
The two V-shapes $y = |2x + 1|$ and $y = |x + 5|$ intersect at $x = -2$ (where $y = 3$) and $x = 4$ (where $y = 9$). So the equation $|2x + 1| = |x + 5|$ has solutions $x = 4$ or $x = -2$.
Since $|a| = |b| \iff a^2 = b^2$, square both sides:
$$(2x + 1)^2 = (x + 5)^2$$
$$4x^2 + 4x + 1 = x^2 + 10x + 25$$
$$3x^2 - 6x - 24 = 0$$
$$x^2 - 2x - 8 = 0$$
$$(x - 4)(x + 2) = 0$$
$$x = 4 \quad \text{or} \quad x = -2$$
$x = 4$: $|2(4)+1| = |9| = 9$, $|4+5| = |9| = 9$. ✓
$x = -2$: $|2(-2)+1| = |-3| = 3$, $|-2+5| = |3| = 3$. ✓
$$x = 4 \quad \text{or} \quad x = -2$$
Solve the inequality $|2x - 3| < 7$.
Currently plotting:
The V-shape $y = |2x - 3|$ has its vertex at $x = \dfrac{3}{2}$ and rises on both sides. It crosses $y = 7$ at $x = -2$ and $x = 5$. The V-shape is below the line between these two crossing points, so the solution is $-2 < x < 5$.
Since $|y| < a \iff -a < y < a$ (for $a > 0$):
$$-7 < 2x - 3 < 7$$
Add 3 to all parts:
$$-4 < 2x < 10$$
Divide by 2:
$$-2 < x < 5$$
$$-2 < x < 5$$
Solve the inequality $2x + 5 < |x + 1|$.
Split into two cases based on the sign of $x + 1$. For each case, replace $|x + 1|$ with the appropriate expression, solve the resulting linear inequality, then check that your solution is consistent with the case's domain.
Case 1: $x + 1 \geq 0$, i.e. $x \geq -1$. Then $|x + 1| = x + 1$.
$$2x + 5 < x + 1 \implies x < -4$$
But $x \geq -1$ and $x < -4$ is impossible. No solution in this case.
Then $|x + 1| = -(x + 1) = -x - 1$.
$$2x + 5 < -x - 1 \implies 3x < -6 \implies x < -2$$
Combined with $x < -1$: $\quad x < -2$.
$x = -3$: $2(-3) + 5 = -1$, $|-3 + 1| = |-2| = 2$. Is $-1 < 2$? ✓
$x = -1.5$: $2(-1.5) + 5 = 2$, $|-1.5 + 1| = |-0.5| = 0.5$. Is $2 < 0.5$? ✗ (not in solution set) ✓
$x = -2$: $2(-2) + 5 = 1$, $|-2 + 1| = |-1| = 1$. Is $1 < 1$? ✗ (boundary, not included) ✓
Sketch $y = 2x + 5$ (a straight line) and $y = |x + 1|$ (V-shape, vertex at $x = -1$). The line $y = 2x + 5$ crosses the $x$-axis at $x = -2.5$ and the V-shape has its vertex at $(-1, 0)$. For $x < -1$, the V-shape is $y = -x - 1$. The line $y = 2x + 5$ is below $y = -x - 1$ when $x < -2$, confirming the solution.
$$x < -2$$
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
Which of the following correctly describes the graph of $y = |x - 2|$?
The expression inside $|\cdot|$ gives the $x$-coordinate of the vertex. For $y = |x - 2|$, the vertex is at $x = 2$, $y = 0$, i.e. $(2, 0)$.
Solve the equation $|x - 3| = |2x + 1|$.
Using $|a| = |b| \iff a^2 = b^2$: $(x - 3)^2 = (2x + 1)^2 \Rightarrow 3x^2 + 10x - 8 = 0 \Rightarrow (3x - 2)(x + 4) = 0$, so $x = \dfrac{2}{3}$ or $x = -4$.
Solve the inequality $|x + 2| < 5$.
Write $|x + 2| < 5$ as $|x - (-2)| < 5$. Using $|x - a| < b \iff a - b < x < a + b$: $-2 - 5 < x < -2 + 5$, i.e. $-7 < x < 3$.
To solve $|x - 1| > 2x$ algebraically, what is the correct first step?
When the modulus is being compared to a variable expression (not a constant), squaring both sides can introduce extraneous solutions. The safe approach is to split into two cases based on the sign of the expression inside the modulus: Case 1 ($x - 1 \geq 0$) and Case 2 ($x - 1 < 0$). Each case is then solved as a linear inequality, and the solution must be checked against the case's domain.
Solve the inequality $|3x - 2| \leqslant 4$.
Using $|y| \leqslant a \iff -a \leqslant y \leqslant a$:
$$-4 \leqslant 3x - 2 \leqslant 4$$
Add 2:
$$-2 \leqslant 3x \leqslant 6$$
Divide by 3:
$$-\dfrac{2}{3} \leqslant x \leqslant 2$$
Solve the equation $|4x - 1| = |9 - x|$.
Using $|a| = |b| \iff a^2 = b^2$, square both sides:
$$(4x - 1)^2 = (9 - x)^2$$
$$16x^2 - 8x + 1 = 81 - 18x + x^2$$
$$15x^2 + 10x - 80 = 0$$
$$3x^2 + 2x - 16 = 0$$
Using the quadratic formula: $x = \dfrac{-2 \pm \sqrt{4 + 192}}{6} = \dfrac{-2 \pm 14}{6}$
$$x = 2 \quad \text{or} \quad x = -\dfrac{8}{3}$$
$x = 2$: $|4(2)-1| = |7| = 7$, $|9-2| = |7| = 7$. ✓
$x = -\dfrac{8}{3}$: $\left|4\left(-\dfrac{8}{3}\right)-1\right| = \left|-\dfrac{35}{3}\right| = \dfrac{35}{3}$, $\left|9-\left(-\dfrac{8}{3}\right)\right| = \left|\dfrac{35}{3}\right| = \dfrac{35}{3}$. ✓
Express the inequality $-1 < x < 3$ in the form $|x - a| < b$, where $a$ and $b$ are to be found.
The midpoint of $-1$ and $3$ is:
$$a = \dfrac{-1 + 3}{2} = 1$$
The half-width is:
$$b = \dfrac{3 - (-1)}{2} = 2$$
$$|x - 1| < 2$$
Solve the inequality $|x + 3| < |x - 4|$.
Both sides have modulus. Square both sides to remove the modulus signs.
Square both sides (valid since both sides are non-negative):
$$(x + 3)^2 < (x - 4)^2$$
$$x^2 + 6x + 9 < x^2 - 8x + 16$$
$$14x < 7$$
$$x < \dfrac{1}{2}$$
$x = 0$: $|3| = 3$, $|-4| = 4$. Is $3 < 4$? ✓
$x = 1$: $|4| = 4$, $|-3| = 3$. Is $4 < 3$? ✗ (not in solution set) ✓
Given that $a$ is a positive constant, solve the inequality $|x - 3a| > |x - a|$.
Square both sides. Remember that $a > 0$, so you can divide by $a$ without flipping the inequality.
Square both sides:
$$(x - 3a)^2 > (x - a)^2$$
$$x^2 - 6ax + 9a^2 > x^2 - 2ax + a^2$$
$$-4ax + 8a^2 > 0$$
$$-4a(x - 2a) > 0$$
Since $a > 0$, we have $-4a < 0$, so divide by $-4a$ and flip the inequality:
$$x - 2a < 0$$
$$x < 2a$$
Exit Ticket
Check the learning objectives you feel confident about: