Edexcel IAL Maths
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Expanding products of brackets and factorising cubics, using f(x) notation
Sketching cubic graphs, including repeated roots, with all axis intercepts
Sketching y=k/x and y=k/x² with their asymptotes, for positive and negative k
Using the points where two graphs meet to state the number of solutions of an equation
The transformations y=af(x), y=f(x)+a, y=f(x+a) and y=f(ax)
The effect of each transformation on intercepts, turning points and asymptotes
Applying the same transformations to the graphs of sin x, cos x and tan x
1. Once a cubic is factorised you can sketch it in seconds, and once you know how each transformation acts you can move that sketch anywhere; a pair of sketches then tells you how many solutions an equation has without solving it.
2. Factorise first: a cubic with no constant term has a common factor x; take it out and factorise the quadratic that is left.
3. Cubic shape: a positive x³ coefficient starts bottom left and ends top right; a negative one does the reverse. A single root crosses the axis; a squared (repeated) root touches it.
4. y=k/x and y=k/x² have asymptotes x=0 and y=0. For k>0, k/x is in the 1st and 3rd quadrants; k/x² is above the x-axis on both sides.
5. The number of solutions of f(x)=g(x) equals the number of points where y=f(x) and y=g(x) meet.
6. Outside the bracket acts on y as you expect: f(x)+a moves up a; af(x) multiplies y by a. Inside the bracket acts on x the opposite way: f(x+a) moves left a; f(ax) divides x by a.
A polynomial is a sum of terms of the form axⁿ, where each power n is a whole number ≥0. Its degree is the highest power: a cubic has degree 3, for example 2x³-3x²-23x+12.
Function notation f(x) names a polynomial so you can substitute into it. If f(x)=2x³-3x²-23x+12, then f(1)=2-3-23+12=-12. This notation is used throughout the course, and it is what makes transformations such as f(x+2) easy to write down.
Expanding three brackets
Multiply two of the brackets first, simplify, then multiply the result by the third. Every term of one bracket must meet every term of the other, so a bracket with two terms times a bracket with three terms gives six products before you collect like terms.
Factorising a cubic in P1
In this unit, cubics to factorise always have a common factor (usually x). Take it out, then factorise the quadratic that remains. Watch for two special patterns:
a difference of two squares, x²-a²=(x-a)(x+a);
a perfect square, such as x²-6x+9=(x-3)².
"Factorise completely" means that no bracket can be factorised any further.
WORKED EXAMPLE · QUESTION
Expand and simplify (x+3)(2x-1)(x-4), giving your answer in the form ax³+bx²+cx+d.
SOLUTION
Multiply two brackets: (x+3)(x-4)=x²-4x+3x-12=x²-x-12.
Multiply by the third bracket, term by term: (2x-1)(x²-x-12)=2x³-2x²-24x-x²+x+12.
Collect like terms: 2x³-3x²-23x+12
Check with x=1: the brackets give 4×1×(-3)=-12, and the answer gives 2-3-23+12=-12 ✓. Calling this f(x), we also know f(-3)=f(4)=f(½)=0, because each value makes one bracket zero.
WORKED EXAMPLE · QUESTION
Factorise completely (a) 3x³-12x (b) x³-6x²+9x (c) 2x³+5x²-3x.
SOLUTION
(a) Common factor 3x, then a difference of two squares: 3x(x²-4)= 3x(x-2)(x+2)
(b) Common factor x, then a perfect square: x(x²-6x+9)= x(x-3)²
(c) Common factor x. For 2x²+5x-3, find two numbers with product 2×(-3)=-6 and sum 5: they are 6 and -1. So x(2x²+6x-x-3)=x(2x(x+3)-(x+3))= x(2x-1)(x+3)
Common error
Stopping at 3x(x²-4). The question says "completely", and x²-4 is a difference of two squares, so the answer is not finished until it is 3x(x-2)(x+2).
A sketch is not a plot. It shows the correct shape and the key points, labelled with their coordinates, but it does not need to be to scale. For a cubic, the factorised form gives you almost everything at once.
What a sketch must show
The right shape. For ax³+… with a>0: as x→∞, y→∞, and as x→-∞, y→-∞. For a<0 the ends are reversed. Find the sign of a by multiplying the x-terms of each bracket.
Where it meets the x-axis: from the factorised form, set y=0.
Where it meets the y-axis: put x=0. For y=(x-p)(x-q)(x-r) this is -pqr.
Repeated roots. A factor (x-p)² means the curve touches the x-axis at x=p, with a turning point on the axis. A factor (x-p)³, as in y=(x-1)³, means the curve flattens and crosses there.
The turning points do not need exact coordinates in a sketch (finding them needs differentiation, P1.6), but a maximum must sit above the axis and a minimum below it wherever the roots force that.
WORKED EXAMPLE · QUESTION
Sketch the curve with equation y=(3-x)(x+1)², showing the coordinates of the points where it meets the axes.
SOLUTION
Shape. The x³ term is (-x)(x²)=-x³: a negative cubic, starting top left and ending bottom right.
x-axis, y=0. x=3 (single root: crosses) and x=-1, a repeated root (touches).
y-axis, x=0. y=3×1²=3, so (0, 3).
Draw it. The curve comes down from the top left, touches the axis at (-1, 0), rises through (0, 3) to a maximum and comes down to cross at (3, 0). Touches at (-1, 0), crosses at (3, 0), meets the y-axis at (0, 3)
Expanded, y=-x³+x²+5x+3: the constant term 3 confirms the y-intercept.
Exam tip
At a touching root, draw the curve genuinely turning on the axis, like the vertex of a parabola: not crossing it and not stopping short of it. Write the coordinates on the sketch itself, not only in your working.
An asymptote is a line that a curve approaches more and more closely, without reaching it, as x or y grows without limit.
| y=k/x | y=k/x² | |
|---|---|---|
| k>0 | two branches, 1st and 3rd quadrants | two branches, both above the x-axis (1st and 2nd quadrants) |
| k<0 | 2nd and 4th quadrants | both below the x-axis (3rd and 4th quadrants) |
| Asymptotes | x=0 and y=0 | x=0 and y=0 |
| Symmetry | rotational, order 2, about O | reflection in the y-axis |
| Meets the axes? | never | never |
Why x=0 is an asymptote: as x→0, the fraction becomes enormous, and x=0 itself gives no value. Why y=0 is an asymptote: as x→±∞ the fraction tends to 0 but never equals it, because k≠0.
Compare the two families: y=k/x² falls away faster than y=k/x for |x|>1 and climbs faster near the y-axis. Both pass through (1, k).
Every point where y=f(x) and y=g(x) meet has an x-coordinate that satisfies f(x)=g(x). So the number of real solutions of f(x)=g(x) equals the number of intersection points, and you can count them without solving the equation.
Questions often hide this link. An equation such as x⁴-4x²-3=0 is really "cubic = reciprocal" after rearranging, and you are expected to show that rearrangement line by line. When you multiply through by x or x², note that x≠0: neither reciprocal curve exists at x=0, so no solutions are lost or gained.
WORKED EXAMPLE · QUESTION
(a) On the same axes, sketch y=x(x²-4) and y=3/x.
(b) Show that the x-coordinates of the points of intersection satisfy x⁴-4x²-3=0.
(c) State the number of real solutions of this equation and find them exactly.
SOLUTION
(a) Cubic: x(x-2)(x+2), positive, roots -2, 0, 2. Reciprocal: 1st and 3rd quadrants, asymptotes the axes. See the figure.
(b) Set the two expressions equal and multiply by x (x≠0): x(x²-4)=3/x⇒x⁴-4x²=3⇒x⁴-4x²-3=0 as required.
(c) The curves meet twice, so there are 2 real solutions. Solve as a quadratic in x²: x²=(4±√(16+12))/2=2±√7.
2-√7<0 is impossible for x², so x=±√(2+√7) (≈±2.16)
The algebra confirms the sketch: one positive and one negative solution, matching Q and P.
Common error
Giving four solutions because a quartic "has four roots". A quartic has at most four real roots; here two values of x² are found, but one is negative, so it gives no real x.
| New graph | Transformation | Point (x, y) moves to |
|---|---|---|
| y=f(x)+a | translation by a units in the positive y-direction (up if a>0) | (x, y+a) |
| y=f(x+a) | translation by a units in the negative x-direction (left if a>0) | (x-a, y) |
| y=af(x) | stretch parallel to the y-axis, scale factor a | (x, ay) |
| y=f(ax) | stretch parallel to the x-axis, scale factor 1/a | (x/a, y) |
Outside the function (f(x)+a, af(x)) the change happens to y, exactly as written. Inside the function (f(x+a), f(ax)) the change happens to x, and it is the inverse: +a moves left, ×a divides by a. The reason: the new graph reaches the old height f(p) when x+a=p, that is at x=p-a, which is a units to the left.
a=-1 gives reflections: -f(x) reflects in the x-axis; f(-x) reflects in the y-axis.
Points on the axis that is not being stretched stay put: x-intercepts are unchanged by af(x); the y-intercept is unchanged by f(ax).
A turning point stays a turning point (a maximum stays a maximum unless a<0 flips it).
WORKED EXAMPLE · QUESTION
The curve y=f(x), where f(x)=x³-3x²-9x, passes through the origin, has a maximum at (-1, 5) and a minimum at (3, -27). Write down the coordinates of the turning points of (a) y=f(x-2) (b) y=3f(x) (c) y=f(½x). (d) Find the number of distinct real roots of f(x)+27=0.
SOLUTION
(a) Translation 2 units in the positive x-direction: maximum (1, 5), minimum (5, -27)
(b) Stretch parallel to the y-axis, scale factor 3: maximum (-1, 15), minimum (3, -81)
(c) Stretch parallel to the x-axis, scale factor 1÷½=2: maximum (-2, 5), minimum (6, -27)
(d) y=f(x)+27 moves the minimum to (3, 0), on the axis, and the maximum to (-1, 32). The curve touches the axis at x=3 and crosses once to the left: 2 distinct roots
Check: x³-3x²-9x+27=(x-3)²(x+3), with roots 3 (repeated) and -3.
Exam tip
To check an "inside the bracket" answer, substitute back: for y=f(x-2) at x=1, you get f(1-2)=f(-1)=5, the old maximum. If the substitution does not land on the old point, the shift went the wrong way.
Asymptotes move with the graph
A vertical asymptote x=c is moved by f(x+a) to x=c-a, and by f(ax) to x=c/a.
A horizontal asymptote y=d is moved by f(x)+a to y=d+a, and by af(x) to y=ad.
An asymptote along an axis that is being stretched towards stays where it is: y=0 is still an asymptote of 5/x, and x=0 is still one of 1/2x.
After moving the asymptotes, find any new intercepts by substitution: a translated reciprocal curve can now cross an axis.
WORKED EXAMPLE · QUESTION
For each curve, state the transformation of y=k/x or y=k/x², the equations of its asymptotes and the coordinates of any points where it meets the axes. (a) y=3/(x+2) (b) y=4/x²-1
SOLUTION
(a) f(x+2) with f(x)=3/x: a translation 2 units in the negative x-direction. Asymptotes x=-2 and y=0. x=0 gives y=(3/2). It never meets the x-axis, since 3 divided by anything is not 0. Asymptotes x=-2, y=0; meets the axes at (0, 1.5) only
(b) g(x)-1 with g(x)=4/x²: a translation 1 unit in the negative y-direction. Asymptotes x=0 and y=-1. y=0 gives x²=4. It never meets the y-axis, since x=0 is excluded. Asymptotes x=0, y=-1; meets the axes at (-2, 0) and (2, 0)
Trig graphs
The same four rules apply to y=sin x, cos x and tan x (in degrees here; radians work in the same way).
asin x: the amplitude becomes |a|.
sin ax: the period 360° is divided by a, so sin 2x has period 180°.
sin(x+a): the whole wave moves a° to the left; each zero, maximum and minimum moves with it.
The asymptotes of tan x at 90° and 270° move too: tan 2x has asymptotes every 90°, starting at 45°.
WORKED EXAMPLE · QUESTION
(a) For 0≤x≤360°, find the coordinates of the points where y=cos(x+60°) meets the axes, and of its maximum and minimum points.
(b) State the period of y=tan 2x and the equations of its asymptotes for 0≤x≤360°.
SOLUTION
(a) y-axis: x=0 gives cos 60°=½, so (0, ½).
x-axis: cos(x+60°)=0 when x+60°=90° or 270° (with 60°≤x+60°≤420°), so (30°, 0) and (210°, 0).
Minimum where x+60°=180°; maximum where x+60°=360°. Minimum (120°, -1), maximum (300°, 1)
(b) A stretch parallel to the x-axis, scale factor ½: the period is 180°÷2. Period 90°; asymptotes x=45°, 135°, 225°, 315°
Common error
Moving cos(x+60°) 60° to the right, or giving sin 2x a period of 720°. Both come from treating changes inside the bracket like changes outside it.
A typical long question walks through the whole topic: factorise, sketch, add a second graph to count solutions, then transform. Each part feeds the next, so a correct sketch early on is worth the time.
WORKED EXAMPLE · QUESTION
f(x)=x³-2x²-3x.
(a) Factorise f(x) completely.
(b) Sketch y=f(x), showing the coordinates of the points where the curve meets the axes.
(c) On the same axes, sketch y=2/x². Hence state, with a reason, the number of real solutions of x⁵-2x⁴-3x³-2=0.
(d) Sketch y=f(x+1), showing the coordinates of the points where it meets the axes.
SOLUTION
(a) Common factor, then the quadratic: two numbers with product -3 and sum -2 are -3 and 1. f(x)=x(x-3)(x+1)
(b) Positive cubic; roots -1, 0, 3; it passes through the origin, so the y-intercept is (0, 0). Crosses at (-1, 0), (0, 0), (3, 0)
(c) 2/x²: two branches above the x-axis, asymptotes x=0 and y=0. Link the equation to the curves (x≠0): x³-2x²-3x=2/x²⇔x⁵-2x⁴-3x³-2=0.
Count the intersections.
For x<-1 and 0<x<3 the cubic is negative and the reciprocal positive: no meeting. For -1<x<0 the cubic stays below 1 (its maximum is about 0.88) while 2/x²>2: no meeting. For x>3 the cubic rises from 0 to ∞ while 2/x² falls towards 0: exactly one meeting. The curves meet once, so there is 1 real solution
(d) Translation 1 unit in the negative x-direction: subtract 1 from each root; the y-intercept is f(0+1)=f(1)=1-2-3=-4. Crosses at (-2, 0), (-1, 0), (2, 0); meets the y-axis at (0, -4)
Exam tip
In a "hence" counting question, the reason is part of the answer. Write a sentence such as "the curves meet at one point, so the equation has one real solution", and show the algebra that turns the given equation into "curve 1 = curve 2".
Exam tip 1 — label everything on the sketch
Write the coordinates of every axis crossing next to the point, and the equation of every asymptote next to its dashed line. A correct shape with no labels shows only part of the answer.
Exam tip 2 — read the ends from the leading term
Before plotting any roots, decide the sign of the x³ term (multiply the x-terms of each bracket). It fixes which corner the curve starts in, and the roots then decide the rest.
Exam tip 3 — inside does the opposite
Changes inside the bracket act on x the opposite way: f(x+3) moves left 3, f(4x) divides x-coordinates by 4. Outside the bracket, the change to y is exactly what you see.
Exam tip 4 — check by substituting one point
After expanding, test x=1 in both forms. After transforming, put your new point into the new equation and confirm it gives the old y-value. Each check takes seconds.
Exam tip 5 — read the command word
"Sketch" means the correct shape with key points labelled, not plotted to scale. "Show that" needs every line of algebra leading to the given result. "Hence" means use the previous part, which here is usually the sketch.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| f(x+2) moves the graph 2 units right. | The old value f(p) is now reached at x=p-2. | f(x+a) is a translation a units to the left. |
| f(2x) doubles every x-coordinate. | The old value f(p) is reached when 2x=p. | f(ax) divides x-coordinates by a: scale factor 1/a. |
| A repeated root is where the curve crosses twice. | A squared factor does not change sign either side of the root. | The curve touches the axis there; it is a turning point. |
| y=3/x² has a branch below the x-axis, like 3/x. | x²>0 for every x≠0. | Both branches are on the same side of the axis as the sign of k. |
| Asymptotes stay on the axes after a transformation. | The asymptote is part of the picture and moves with it. | Apply the same transformation to the asymptote's equation. |
| A stretch af(x) moves the roots. | 0×a=0, so points on the x-axis stay there. | Roots are unchanged by af(x); only the y-coordinates are scaled. |
P1.1 Quadratics and P1.6 Differentiation
The discriminant from P1.1 decides whether a line meets a curve; differentiation finds the exact turning points that a sketch here only shows roughly.
Factor theorem and algebraic division (P2); functions (P3)
When a cubic has no common factor of x, you find one root by trying values in f(x) and divide, then sketch exactly as here. Two or more transformations in sequence, modulus graphs |f(x)| and f(|x|), and inverse functions in P3 all build on the four rules on this page.
IAL Physics and Mechanics
Inverse-square laws (gravitational and electric field strength) follow y=k/x², and Boyle's law, p=k/V, is a y=k/x graph.
| Formula | Use | In the formulae list? |
|---|---|---|
| x²-a²=(x-a)(x+a) | difference of two squares when factorising | No – learn it |
| y=(x-p)(x-q)(x-r) | cubic with roots p, q, r; y-intercept -pqr | No – learn it |
| y=k/x, y=k/x² | reciprocal graphs; asymptotes x=0 and y=0 | No – learn it |
| y=f(x)+a | translation a units in the positive y-direction | No – learn it |
| y=f(x+a) | translation a units in the negative x-direction | No – learn it |
| y=af(x) | stretch parallel to the y-axis, scale factor a | No – learn it |
| y=f(ax) | stretch parallel to the x-axis, scale factor 1/a | No – learn it |
1. Expand (x-2)³.
2. State the equations of the asymptotes of y=5/(x-4).
3. The point (6, -2) lies on y=f(x). Write down its image on y=f(3x) and on y=f(x)-5.
4. How many times does y=x³ meet y=-1/x?
5. Write down the coordinates of the maximum and minimum points of y=4sin x for 0≤x≤360°.
6. Factorise x³-x²-12x completely and state where the curve y=x³-x²-12x meets the x-axis.
1. (x-2)²=x²-4x+4; multiplying by (x-2) gives x³-4x²+4x-2x²+8x-8= x³-6x²+12x-8.
2. It is 5/x translated 4 units in the positive x-direction: x=4 and y=0.
3. f(3x): divide the x-coordinate by 3, giving (2, -2). f(x)-5: subtract 5 from the y-coordinate, giving (6, -7).
4. Never. For x>0, x³ is positive and -1/x negative; for x<0 it is the other way round. The algebra agrees: x⁴=-1 has no real solution.
5. A stretch parallel to the y-axis, scale factor 4: maximum (90°, 4), minimum (270°, -4).
6. x(x²-x-12); two numbers with product -12 and sum -1 are -4 and 3, so x(x-4)(x+3). The curve crosses the x-axis at (-3, 0), (0, 0) and (4, 0).
Check understanding
The answer and each step appear one click at a time.
A cubic curve is given in factorised form. Its equation is y = (x − 4)(x + 1)(x + 3), and the curve is labelled C.
(a) Sketch C, showing the coordinates of the points where C meets the coordinate axes. [3]
(b) Show that the equation of C can be written as y = x³ − 13x − 12. [2]
(c) The line y = −12 meets C where C crosses the y-axis and at two other points. Find the exact x-coordinates of these two points. [2]
Mark scheme click to reveal each step
positive cubic shape: from bottom left to top right with a maximum and a minimum
crosses the x-axis at (−3, 0), (−1, 0) and (4, 0)
meets the y-axis at (0, −12)
(x + 1)(x + 3) = x² + 4x + 3
(x − 4)(x² + 4x + 3) = x³ + 4x² + 3x − 4x² − 16x − 12 = x³ − 13x − 12
x³ − 13x − 12 = −12, so x³ − 13x = 0, x(x² − 13) = 0
x = √13 and x = −√13
The curve with equation y = f(x) passes through the origin O, has a maximum point at A(2, 4) and meets the x-axis again at B(5, 0). The curve has no other turning points.
Figure: sketch of y = f(x)
(a) Write down the coordinates of the images of O, A and B on the curve y = f(x) − 4. [2]
(b) Write down the coordinates of the images of O, A and B on the curve y = f(x + 2). [2]
(c) Write down the coordinates of the maximum point of y = 3f(x). [1]
(d) Write down the coordinates of the images of A and B on the curve y = f(2x). [1]
(e) State, with a reason, the number of solutions of the equation f(x) = 4. [1]
Mark scheme click to reveal each step
maximum at (2, 0)
(0, −4) and (5, −4)
maximum at (0, 4)
(−2, 0) and (3, 0)
(2, 12)
(1, 4) and (2.5, 0)
one solution, because the line y = 4 only touches the curve at its maximum A, the only turning point (accept: f(x) − 4 = 0 has its maximum on the x-axis, as in part (a))
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