The modulus of a number is its size without its sign. Apply it to a whole function and every negative output becomes positive, so the graph of \(y = |f(x)|\) is the graph of \(y = f(x)\) with everything below the \(x\)-axis reflected above it. Learn to sketch it, find its key points and use it to count solutions.
\(|f(x)| = f(x)\) where \(f(x) \ge 0\) and \(|f(x)| = -f(x)\) where \(f(x) < 0\). So to draw \(y = |f(x)|\), keep the parts of \(y = f(x)\) on or above the \(x\)-axis and reflect the parts below it in the \(x\)-axis. The graph is never below the \(x\)-axis, and it has a sharp corner wherever \(y = f(x)\) crosses the axis.
Fig. 1: y = f(x) (dashed) and y = |f(x)| (solid) for f(x) = x² - 2x - 3. The part below the x-axis is reflected, the minimum (1, -4) becomes a maximum (1, 4), and the graph has sharp corners at x = -1 and x = 3
Section 01
Reflect what lies below.
The modulus \(|a|\) is the distance of \(a\) from zero, so \(|3| = 3\) and \(|-3| = 3\). For a function, \(|f(x)|\) leaves every positive or zero output alone and changes the sign of every negative output. On a graph, changing the sign of \(y\) is a reflection in the \(x\)-axis, but it happens ONLY where the curve is below the axis.
Step 1. Sketch \(y = f(x)\) and find where it meets the \(x\)-axis (solve \(f(x) = 0\)).
Step 2. Keep every part on or above the \(x\)-axis unchanged.
Step 3. Reflect every part below the \(x\)-axis in the \(x\)-axis.
Step 4. Label the \(x\)-intercepts (unchanged), the \(y\)-intercept \((0, |f(0)|)\) and any minimum that has become a maximum.
Where the curve crosses the axis at an angle, the reflected piece meets the kept piece in a sharp corner (a V point), not a smooth turn. Fig. 1 shows the whole process for a quadratic.
Worked example A linear function
Sketch \(y = |2x - 4|\), stating the coordinates of the points where the graph meets the axes.
\(y = 2x - 4\) is a straight line through \((2, 0)\) and \((0, -4)\), with gradient \(2\).
It is below the \(x\)-axis for \(x < 2\). Reflect that half: the line becomes \(y = -(2x - 4) = 4 - 2x\) there.
The \(y\)-intercept \((0, -4)\) moves to \((0, 4)\); the \(x\)-intercept \((2, 0)\) stays.
The graph is a V shape with its sharp corner at \((2, 0)\).
Answer: A V shape with vertex \((2, 0)\), meeting the \(y\)-axis at \((0, 4)\)
Worked example A quadratic function (Fig. 1)
Sketch \(y = |x^{2} - 2x - 3|\). State the coordinates of the points where it meets the axes and of its local maximum.
\(x^{2} - 2x - 3 = (x + 1)(x - 3)\), so \(y = f(x)\) meets the \(x\)-axis at \(x = -1\) and \(x = 3\).
Complete the square: \(x^{2} - 2x - 3 = (x - 1)^{2} - 4\), so the minimum of \(y = f(x)\) is \((1, -4)\).
The curve is below the axis for \(-1 < x < 3\). Reflecting this arc turns the minimum \((1, -4)\) into a local maximum \((1, 4)\).
\(f(0) = -3\), so the \(y\)-intercept of \(y = |f(x)|\) is \((0, 3)\). There are sharp corners at \((-1, 0)\) and \((3, 0)\).
Answer: Meets the axes at \((-1, 0)\), \((3, 0)\) and \((0, 3)\); local maximum \((1, 4)\)
Your turn
Write down the coordinates of the points where \(y = |3x + 6|\) meets the axes.
\(3x + 6 = 0\) at \(x = -2\), so \((-2, 0)\), the sharp corner. \(|3(0) + 6| = 6\), so \((0, 6)\).
Examiner trap
Only the parts BELOW the \(x\)-axis move. Do not reflect the whole curve, and do not reflect in the \(y\)-axis: \(y = |f(x)|\) is not the same as \(y = f(|x|)\). Check your sketch: no part of \(y = |f(x)|\) may lie below the \(x\)-axis.
Section 02
Cubic and trigonometric curves.
The method is the same for any function: find where \(f(x) = 0\), decide where \(f(x)\) is negative, and reflect those pieces. For a cubic, the factors tell you the sign in each interval. For \(y = a\sin bx + c\), \(y = a\cos bx + c\) or \(y = a\tan bx + c\), sketch the curve over the given domain first, then reflect.
Range of \(|f(x)|\). If \(f(x)\) takes values from \(m\) to \(M\) on a domain and \(m < 0 < M\), then \(|f(x)|\) takes every value from \(0\) up to the larger of \(|m|\) and \(M\). The least value is \(0\) (the curve touches the axis), not \(|m|\). If \(f(x)\) never changes sign, just make both ends of its range positive.
Worked example A cubic
For \(f(x) = (x + 1)(x - 1)(x - 3)\), state the intervals in which the graph of \(y = f(x)\) is reflected to give \(y = |f(x)|\), and find the coordinates of the points where \(y = |f(x)|\) meets the axes.
\(f(x) = 0\) at \(x = -1\), \(1\) and \(3\). The coefficient of \(x^{3}\) is positive, so \(f(x)\) is negative for very negative \(x\).
So the pieces for \(x < -1\) and \(1 < x < 3\) are reflected in the \(x\)-axis.
\(f(0) = (1)(-1)(-3) = 3 > 0\), so the \(y\)-intercept \((0, 3)\) is unchanged.
Answer: Reflect for \(x < -1\) and \(1 < x < 3\); meets the axes at \((-1, 0)\), \((1, 0)\), \((3, 0)\) (all sharp corners) and \((0, 3)\)
Worked example A cosine curve
Sketch \(y = |2\cos x + 1|\) for \(0^{\circ} \le x \le 360^{\circ}\) and state its range.
\(y = 2\cos x + 1\) starts at \(3\) when \(x = 0^{\circ}\), falls to its minimum \(-1\) at \(x = 180^{\circ}\) and returns to \(3\) at \(x = 360^{\circ}\).
It meets the \(x\)-axis where \(\cos x = -\dfrac{1}{2}\): \(x = 120^{\circ}\) and \(x = 240^{\circ}\).
For \(120^{\circ} < x < 240^{\circ}\) it is negative, so reflect that piece: the minimum \((180^{\circ}, -1)\) becomes a local maximum \((180^{\circ}, 1)\).
The values of \(2\cos x + 1\) run from \(-1\) to \(3\), so \(|2\cos x + 1|\) runs from \(0\) to the larger of \(1\) and \(3\).
Answer: Range \(0 \le y \le 3\); sharp corners at \(120^{\circ}\) and \(240^{\circ}\), local maximum \((180^{\circ}, 1)\)
Your turn
\(f(x) = 2x - 5\) for \(0 \le x \le 4\). Find the range of \(f\) and the range of \(|f|\).
\(f(0) = -5\) and \(f(4) = 3\), so \(-5 \le f(x) \le 3\). The line crosses zero at \(x = 2.5\), so \(0 \le |f(x)| \le 5\).
Examiner trap
The range of \(|f(x)|\) is NOT found by taking the modulus of the two ends. For \(-5 \le f(x) \le 3\), the answer \(3 \le |f(x)| \le 5\) is wrong: \(f\) passes through \(0\), so the least value of \(|f(x)|\) is \(0\).
Stage 1 · Learn
Check your understanding
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To obtain the graph of \(y = |f(x)|\) from the graph of \(y = f(x)\), which parts are reflected in the \(x\)-axis?
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Where does the graph of \(y = |x^{2} - 5x - 6|\) meet the \(y\)-axis?
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Where does the graph of \(y = |3x - 6|\) have a sharp corner?
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On a certain domain, \(-3 \le f(x) \le 2\) and \(f\) is continuous. What is the range of \(|f(x)|\)?
Section 03
Counting solutions with a horizontal line.
The solutions of \(|f(x)| = k\) are the \(x\)-coordinates of the points where the line \(y = k\) meets the graph of \(y = |f(x)|\). So sketch \(y = |f(x)|\), slide a horizontal line up from the \(x\)-axis and count the intersections. Pay special attention when the line passes through a corner, a local maximum or an end point of the domain. (Solving modulus equations and inequalities algebraically comes later, in Topic 4.)
Worked example Every value of k
Using Fig. 1, state the number of solutions of \(|x^{2} - 2x - 3| = k\) for each value or range of values of \(k\).
\(k < 0\): the line is below the \(x\)-axis, so there are no solutions.
\(k = 0\): the line is the \(x\)-axis and meets the graph only at the corners, giving 2 solutions (\(x = -1\) and \(x = 3\)).
\(0 < k < 4\): the line cuts both outer arms and the reflected arch twice, giving 4 solutions.
\(k = 4\): the line touches the local maximum \((1, 4)\) and cuts the two arms, giving 3 solutions.
\(k > 4\): it is above the arch and cuts only the two arms, giving 2 solutions.
Find the number of solutions of \(|2\cos x + 1| = 1\) for \(0^{\circ} \le x \le 360^{\circ}\).
From the previous section, the graph falls from \(3\) to \(0\) at \(120^{\circ}\), rises to the local maximum \(1\) at \(180^{\circ}\), falls to \(0\) at \(240^{\circ}\) and rises to \(3\) at \(360^{\circ}\).
The line \(y = 1\) cuts the first falling piece once, touches the local maximum at \(180^{\circ}\), and cuts the last rising piece once.
Check: \(2\cos x + 1 = 1\) gives \(\cos x = 0\), so \(x = 90^{\circ}, 270^{\circ}\); \(2\cos x + 1 = -1\) gives \(\cos x = -1\), so \(x = 180^{\circ}\).
How many solutions does \(|x^{2} - 2x - 3| = 2\) have?
\(0 < 2 < 4\), so the line \(y = 2\) cuts the graph in Fig. 1 at 4 points: 4 solutions.
Examiner trap
Count from the graph of \(y = |f(x)|\), not from \(y = f(x)\). The equation \(x^{2} - 2x - 3 = 2\) has only 2 solutions, but \(|x^{2} - 2x - 3| = 2\) has 4, because the reflected arch adds two more. Also check end points: a line through an end of the domain counts that end point as a solution.
Stage 2 · Exam
Exam-style questions
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Find the number of solutions of the equation \(|x^{2} - 4x| = 3\).
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Find the value of \(k\) for which the equation \(|x^{2} - 4x| = k\) has exactly 3 solutions.
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Find the range of \(y = |3\sin 2x - 1|\) for \(0^{\circ} \le x \le 180^{\circ}\).
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The graph of \(y = |(x - 1)(x - 5)|\) has a local maximum. Find its coordinates.
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