When a cubic is given as three brackets, you can sketch it without plotting a single table of values: the brackets give the roots, \(x = 0\) gives the \(y\)-intercept, and the sign of the \(x^{3}\) term says which way the curve starts and ends. Reflect the parts below the \(x\)-axis and you have \(y = |f(x)|\) as well.
For \(y = f(x)\) written as three linear factors: the roots come from setting each factor to zero; the \(y\)-intercept is \(f(0)\); the \(x^{3}\) coefficient is the product of the \(x\)-coefficients of the factors, and if it is positive the curve starts bottom left and ends top right, if negative it starts top left and ends bottom right. A squared factor \((x - a)^{2}\) means the curve touches the \(x\)-axis at \(x = a\) and turns back. For \(y = |f(x)|\), keep every part on or above the \(x\)-axis and reflect every part below it in the \(x\)-axis.
Fig. 1: y = (x + 1)(x - 2)(x - 4). The x³ coefficient is positive, so the curve starts bottom left and ends top right; it crosses the x-axis at the three roots and meets the y-axis at (0, 8)
Section 01
Read the shape from the factors.
A sketch is not an accurate plot. It must have the right shape and every point where the curve meets the axes labelled with its coordinates. For a cubic given as a product of three linear factors, three quick facts give you all of that. You do not need the exact turning points (that needs calculus, which comes later in the course), but the curve must turn once between neighbouring roots and be smooth, with no corners.
Ends: multiply the \(x\)-coefficients of the three brackets to get the \(x^{3}\) coefficient. Positive: the curve starts bottom left and ends top right. Negative: it starts top left and ends bottom right. A bracket written as \((2 - x)\) has \(x\)-coefficient \(-1\), so it flips the ends.
Roots: set each bracket to zero. \((x + 1)\) gives \(x = -1\); \((2x - 1)\) gives \(x = \frac{1}{2}\); \((2 - x)\) gives \(x = 2\).
\(y\)-intercept: put \(x = 0\) into the brackets and multiply.
Draw: start from the correct end, pass through the roots in order from left to right, turning once between each pair, and finish at the other end.
Worked example Positive x³ coefficient
Sketch the curve \(y = (x + 1)(x - 2)(x - 4)\), showing the coordinates of the points where it meets the axes.
Ends: the \(x^{3}\) coefficient is \(1 \times 1 \times 1 = 1 > 0\), so the curve starts bottom left and ends top right.
Roots: \(x = -1\), \(x = 2\) and \(x = 4\), so it crosses the \(x\)-axis at \((-1, 0)\), \((2, 0)\) and \((4, 0)\).
\(y\)-intercept: \(x = 0\) gives \((1)(-2)(-4) = 8\), so it meets the \(y\)-axis at \((0, 8)\).
Draw: up from bottom left through \((-1, 0)\), over a maximum above the axis (passing through \((0, 8)\)), down through \((2, 0)\), round a minimum below the axis, up through \((4, 0)\) and on to top right (Fig. 1).
Answer: An N-shaped curve through \((-1, 0)\), \((0, 8)\), \((2, 0)\) and \((4, 0)\), as in Fig. 1
Worked example A bracket that flips the ends
Sketch \(y = (2x - 1)(x + 3)(2 - x)\), showing where it meets the axes.
Ends: the \(x\)-coefficients are \(2\), \(1\) and \(-1\), so the \(x^{3}\) coefficient is \(2 \times 1 \times (-1) = -2 < 0\). The curve starts top left and ends bottom right.
\(y\)-intercept: \((-1)(3)(2) = -6\), so \((0, -6)\).
Draw: down from top left through \((-3, 0)\), round a minimum below the axis (passing through \((0, -6)\)), up through \((\frac{1}{2}, 0)\), over a maximum, then down through \((2, 0)\) to bottom right.
Check one region: at \(x = 1\), \(y = (1)(4)(1) = 4 > 0\), so the curve is above the axis between \(\frac{1}{2}\) and \(2\), as drawn.
Answer: A curve from top left to bottom right through \((-3, 0)\), \((0, -6)\), \((\frac{1}{2}, 0)\) and \((2, 0)\)
Your turn
State the end behaviour of \(y = x(x + 2)(x - 3)\) and the coordinates of the points where it meets the axes.
The \(x^{3}\) coefficient is \(1 > 0\): bottom left to top right. It crosses the \(x\)-axis at \((-2, 0)\), \((0, 0)\) and \((3, 0)\); the \(y\)-intercept is the origin, since \(0 \times 2 \times (-3) = 0\).
Examiner trap
The root of a bracket has the OPPOSITE sign to the number inside: \((x + 3)\) gives \(x = -3\), not \(3\). A bracket \((2x - 1)\) gives \(x = \frac{1}{2}\), not \(1\) or \(2\). And look for brackets written the 'wrong way round', such as \((2 - x)\): they make the \(x^{3}\) coefficient negative, so the whole curve starts top left.
Section 02
Repeated factors touch, not cross.
If two of the three factors are the same, the cubic has a squared factor \((x - a)^{2}\). A square is never negative, so near \(x = a\) the curve stays on one side of the axis: it comes down (or up) to the axis, touches it at \((a, 0)\) and turns back. That point is a turning point on the \(x\)-axis. The single factor still gives an ordinary crossing.
Fig. 2: y = (x + 2)²(1 - x). The squared factor makes the curve touch the x-axis at (-2, 0); the single factor (1 - x) gives a crossing at (1, 0). The x³ coefficient is negative, so the curve starts top left
\((x - a)^{2}\): the curve touches the \(x\)-axis at \((a, 0)\) and turns back. Only two points on the \(x\)-axis are labelled, not three.
\((x - a)^{3}\), as in \(y = (x - 1)^{3}\): the curve crosses at \((a, 0)\) but flattens out as it passes through, like \(y = x^{3}\) at the origin.
The ends and the \(y\)-intercept are found exactly as before.
Worked example A squared factor and a negative x³ coefficient
Sketch \(y = (x + 2)^{2}(1 - x)\), showing where it meets the axes.
Ends: \((x + 2)^{2}(1 - x)\) has \(x^{3}\) coefficient \(1 \times 1 \times (-1) = -1 < 0\), so top left to bottom right.
Roots: \((x + 2)^{2}\) gives \(x = -2\), a repeated root, so the curve touches the axis at \((-2, 0)\). \((1 - x)\) gives \(x = 1\), a crossing.
\(y\)-intercept: \((2)^{2}(1) = 4\), so \((0, 4)\).
Draw: down from top left to touch the axis at \((-2, 0)\), back up over a maximum (here it happens to be at \((0, 4)\)), then down through \((1, 0)\) to bottom right (Fig. 2).
Answer: Touches at \((-2, 0)\), crosses at \((1, 0)\), meets the \(y\)-axis at \((0, 4)\), as in Fig. 2
Worked example Equation from a sketch
A cubic curve touches the \(x\)-axis at \((-1, 0)\), crosses it at \((3, 0)\) and meets the \(y\)-axis at \((0, -6)\). Find its equation in the form \(y = k(x + 1)^{2}(x - 3)\).
Touching at \(x = -1\) needs the squared factor \((x + 1)^{2}\); crossing at \(x = 3\) needs \((x - 3)\). So \(y = k(x + 1)^{2}(x - 3)\).
Use \((0, -6)\): \(-6 = k(1)^{2}(-3) = -3k\), so \(k = 2\).
Check the shape: the \(x^{3}\) coefficient is \(2 > 0\), so the curve starts bottom left, rises to touch the axis at \((-1, 0)\) from below, falls through \((0, -6)\), turns and rises through \((3, 0)\).
Answer: \(y = 2(x + 1)^{2}(x - 3)\)
Your turn
Describe the sketch of \(y = (3 - x)(x - 1)^{2}\).
The \(x^{3}\) coefficient is \(-1 \times 1 \times 1 = -1\): top left to bottom right. It touches the \(x\)-axis at \((1, 0)\), crosses at \((3, 0)\), and meets the \(y\)-axis at \((0, 3)\), since \((3)(-1)^{2} = 3\).
Examiner trap
A squared factor is ONE point on the axis, not two: \((x - 1)^{2}\) does not mean roots at \(1\) and \(-1\). And the curve must not cross at a repeated root; drawing it straight through \((a, 0)\) loses the shape mark.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Where does \(y = (x - 3)(x + 2)(2x - 1)\) meet the \(y\)-axis?
Quick check+10 XP
Which describes the ends of \(y = (x + 1)(x - 2)(3 - x)\)?
Quick check+10 XP
What happens at \(x = 1\) on the curve \(y = (x - 1)^{2}(x + 4)\)?
Quick check+10 XP
To sketch \(y = |(x + 1)(x - 2)(x - 4)|\), which parts of \(y = (x + 1)(x - 2)(x - 4)\) are reflected in the \(x\)-axis?
Section 03
The modulus graph: reflect what is below.
You met \(y = |f(x)|\) earlier in the course: \(|f(x)|\) is never negative, so every part of \(y = f(x)\) below the \(x\)-axis is reflected in the \(x\)-axis and every other part stays where it is. For a cubic, sketch \(y = f(x)\) first (lightly or dashed), then reflect. The points on the \(x\)-axis stay put; the \(y\)-intercept becomes \(|f(0)|\).
Fig. 3: y = |(x + 1)(x - 2)(x - 4)|. The parts of Fig. 1 below the x-axis (dashed) are reflected, giving sharp V-points at the roots and two arms that both rise upwards. The line y = 6 meets the graph at 4 points
At a crossing root the modulus graph has a sharp V-point on the axis. Do not round it off.
At a touching root the curve is already on one side, so after reflection it still touches smoothly.
Both ends of \(y = |f(x)|\) point upwards, because the end that went down to \(-\infty\) is reflected.
Solutions of \(|f(x)| = k\) are where the horizontal line \(y = k\) meets the graph. For \(k < 0\) there are none; for \(k = 0\) they are the roots; for \(k > 0\) compare \(k\) with the heights of the humps, which a question gives you when it needs them.
Worked example Counting solutions with a horizontal line
\(f(x) = (x + 1)(x - 2)(x - 4)\). The turning points of \(y = f(x)\) are at about \((0.2, 8.2)\) and \((3.1, -4.1)\). Sketch \(y = |f(x)|\) and find the number of solutions of \(|f(x)| = k\) for \(k = 2\), \(k = 6\) and \(k = 10\).
From Fig. 1, \(f(x) < 0\) for \(x < -1\) and for \(2 < x < 4\). Reflect those parts: the left arm now rises from \((-1, 0)\) to top left, and the dip between \(2\) and \(4\) becomes a hump of height \(4.1\) (Fig. 3).
Label \((-1, 0)\), \((2, 0)\), \((4, 0)\) and \((0, 8)\), since \(|f(0)| = 8\).
The graph has two arms going up for ever, a hump of height \(8.2\) on \(-1 < x < 2\) and a hump of height \(4.1\) on \(2 < x < 4\).
\(k = 2\): below both humps, so the line meets each arm once and each hump twice: \(1 + 2 + 2 + 1 = 6\).
\(k = 6\): above the small hump but below the big one: \(1 + 2 + 0 + 1 = 4\) (Fig. 3).
\(k = 10\): above both humps, so only the two arms: \(2\).
Answer: 6 solutions for \(k = 2\), 4 for \(k = 6\), 2 for \(k = 10\)
Worked example A repeated root and an exact turning point
\(f(x) = (x + 2)^{2}(1 - x)\) has a maximum point at \((0, 4)\) (Fig. 2). Sketch \(y = |f(x)|\). Find the number of solutions of \(|f(x)| = 4\), and the values of \(k\) for which \(|f(x)| = k\) has exactly two solutions.
\(f(x) < 0\) only for \(x > 1\), so only the right-hand arm is reflected: it now rises from \((1, 0)\) to top right, with a V-point at \((1, 0)\).
At \((-2, 0)\) the curve already touches the axis from above, so it stays a smooth touch. The \(y\)-intercept is still \((0, 4)\).
\(|f(x)| = 4\): the line meets the left arm once, touches the top of the hump at \(x = 0\), and meets the right arm once: 3 solutions.
For \(0 < k < 4\) there are \(1 + 2 + 1 = 4\) solutions; for \(k > 4\) only the two arms, so 2.
For \(k = 0\) the solutions are the roots \(x = -2\) and \(x = 1\): also exactly 2.
Answer: 3 solutions when \(k = 4\); exactly two solutions when \(k > 4\) or \(k = 0\)
Your turn
For \(y = |x(x + 2)(x - 3)|\), which parts of \(y = x(x + 2)(x - 3)\) are reflected, and where does the graph meet the axes?
\(x(x + 2)(x - 3)\) is negative for \(x < -2\) (three negative factors) and for \(0 < x < 3\) (one negative factor), so those parts are reflected. The graph meets the axes at \((-2, 0)\), \((0, 0)\) and \((3, 0)\) only.
Examiner trap
Reflect only the parts BELOW the axis, and never draw any of \(y = |f(x)|\) below the axis. Keep the V-points sharp at crossing roots, and write the \(y\)-intercept as \(|f(0)|\): for \(y = |(2x - 1)(x + 3)(2 - x)|\) it is \((0, 6)\), not \((0, -6)\). In the next lesson the same sketches are used to solve cubic inequalities such as \(f(x) > 0\).
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
A cubic curve crosses the \(x\)-axis at \(x = -2\), \(x = 1\) and \(x = 3\), and meets the \(y\)-axis at \((0, -6)\). Which could be its equation?
Exam style+20 XP
Which statement describes the curve \(y = (1 - 2x)(x + 3)^{2}\)?
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Sketch \(y = (x - 2)^{2}(x + 1)\). Which points should be labelled on the axes?
Exam style+20 XP
The curve \(y = (x + 1)(x - 2)^{2}\) has a maximum point at \((0, 4)\). Find the set of values of \(k\) for which \(|(x + 1)(x - 2)^{2}| = k\) has exactly four solutions.
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