Edexcel International A Level Maths · P1 · Pure Mathematics 1
Polynomials, curve sketching and transformations
28 free, high-quality past-paper-style questions on Polynomials, curve sketching and transformations for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 246 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P1-3-B-01P1 · ASEasy6 marks
Three linear factors are multiplied together to give the cubic polynomial f(x) = (x − 5)(x + 1)(3x + 2).
(a)
Expand f(x), giving your answer in the form ax3 + bx2 + cx + d, where a, b, c and d are integers.
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)
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Answer
Mark
Notes
Part (a)
maximum at (2, 0)
B1
(0, −4) and (5, −4)
B1
Part (b)
maximum at (0, 4)
B1
(−2, 0) and (3, 0)
B1
Part (c)
(2, 12)
B1
Part (d)
(1, 4) and (2.5, 0)
B1
Part (e)
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))
B1
Common error: Moving the curve y = f(x + 2) two units to the right. A change inside the bracket acts on x in the opposite direction: subtract 2 from each x-coordinate.
Figure 1 shows the curve with equation y = f(x), where f(x) is a cubic polynomial. The curve crosses the x-axis at the three points marked and meets the y-axis at (0, −10).
(a)
Using Figure 1, write f(x) in the form a(x − p)(x − q)(x − r), stating the values of the constants a, p, q and r.
(3)
(b)
Hence express f(x) in the form bx3 + cx2 + dx + e.
(2)
(c)
Using Figure 1, find the set of values of x for which f(x) > 0.
(2)
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Answer
Mark
Notes
Part (a)
roots read from the figure: −2, 2 and 5, so f(x) = a(x + 2)(x − 2)(x − 5)
the curve is above the x-axis to the left of −2 and between 2 and 5
M1
x < −2 or 2 < x < 5
A1
Common error: Taking a = 1 without checking. The y-intercept fixes a, and its sign must match the shape: this curve falls to the right, so a is negative.
Figure 1 shows the curve with equation y = k/x, where k is a constant, and the straight line L. The curve passes through the point A(4, 3). The line L crosses the coordinate axes at the two points marked and meets the curve at the points P and Q.
(a)
Find the value of k.
(1)
(b)
Using Figure 1, find an equation of L.
(1)
(c)
Write down the equations of the asymptotes of the curve y = k/x.
(1)
(d)
Find the coordinates of P and of Q.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
3 = k/4, so k = 12
B1
Part (b)
through (−1, 0) and (0, 1): gradient 1, so y = x + 1
B1
Part (c)
x = 0 and y = 0
B1
Part (d)
12/x = x + 1, so x2 + x − 12 = 0
M1
(x + 4)(x − 3) = 0, x = 3 or x = −4
A1
P(3, 4) and Q(−4, −3)
A1
Exam tip: Use the figure to check the algebra: P is in the first quadrant and Q in the third, so one point must have both coordinates positive and the other both negative.
Figure 1 shows the graph of y = a cos bx for 0 ≤ x ≤ 360°, where a and b are positive constants.
(a)
Using Figure 1, write down the value of a and find the value of b.
(2)
(b)
Describe fully the two stretches that map the graph of y = cos x onto the graph in Figure 1.
(2)
(c)
State the number of solutions of the equation a cos bx = 0.5 for 0 ≤ x ≤ 360°, giving a reason.
(2)
(d)
The graph in Figure 1 is translated by c units in the positive y-direction, where c > 0, so that the new graph touches the x-axis. Write down the value of c and the values of x at which the new graph touches the x-axis.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
greatest value 2, so a = 2
B1
one cycle takes 120°, so b = 360 ÷ 120: b = 3
B1
Part (b)
stretch parallel to the y-axis, scale factor 2
B1
stretch parallel to the x-axis, scale factor ⅓
B1
Part (c)
the line y = 0.5 meets the curve twice in each cycle
M1
there are three complete cycles, so 6 solutions
A1
Part (d)
the minimum value −2 must move up to 0: c = 2
B1
x = 60°, 180°, 300°
B1
Exam tip: Read the period from the figure, not just the first turning point: here the curve repeats every 120°, and b = 360° ÷ period.
Figure 1 shows a rectangular sheet of card measuring 20 cm by 12 cm. A square of side x cm is cut from each corner and the four sides are folded up along the dashed lines to make an open box with a rectangular base. The volume of the box is V cm3.
(a)
Show that V = 4x3 − 64x2 + 240x.
(3)
(b)
Explain why 0 < x < 6.
(1)
(c)
Factorise 4x3 − 64x2 + 240x completely. Hence sketch the curve y = 4x3 − 64x2 + 240x for all real values of x, showing the coordinates of the points where it meets the coordinate axes.
(3)
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Answer
Mark
Notes
Part (a)
base (20 − 2x) by (12 − 2x), height x
B1
(20 − 2x)(12 − 2x) = 240 − 64x + 4x2
M1
V = x(240 − 64x + 4x2) = 4x3 − 64x2 + 240x
A1
Part (b)
x is a length so x > 0, and the width 12 − 2x must be positive, so x < 6
B1
Part (c)
4x(x − 6)(x − 10)
B1
positive cubic shape: from bottom left to top right with a maximum and a minimum
B1
passes through (0, 0) and crosses the x-axis at (6, 0) and (10, 0)
B1
Common error: Writing the base as (20 − x) by (12 − x). A square is cut from both ends of each edge, so each length loses 2x.
This question compares a cubic curve with a reciprocal curve. The cubic, C, has a repeated factor: its equation is y = (x − 1)(x + 2)2.
(a)
Sketch C, showing the coordinates of the points where C meets the coordinate axes.
(3)
(b)
On the same axes, sketch the curve y = 4/x.
(2)
(c)
Hence state, with a reason, the number of real solutions of the equation x(x − 1)(x + 2)2 = 4.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
positive cubic shape
B1
touches the x-axis at (−2, 0) and crosses it at (1, 0)
B1
meets the y-axis at (0, −4)
B1
Part (b)
two branches, in the first and third quadrants
B1
approaching both coordinate axes, which are its asymptotes, and not meeting them
B1
Part (c)
dividing by x (x ≠ 0) gives (x − 1)(x + 2)2 = 4/x, so the solutions are where the two curves meet
M1
the curves meet twice (once where x > 1 and once where x < −2), so there are 2 real solutions
A1
Model sketch
Figure: model sketch of C and y = 4/x
Exam tip: For a “hence” counting question, write the algebra that turns the given equation into “curve = curve” before you count. The count alone does not show where it came from.
The function f is defined by f(x) = (x + 1)(x − 3)2. In parts (b) and (c) its graph is moved by a translation and by a stretch.
(a)
Sketch the curve y = f(x), showing the coordinates of the points where it meets the coordinate axes.
(3)
(b)
Sketch the curve y = f(x − 2), showing the coordinates of the points where it meets the coordinate axes.
(3)
(c)
The curve y = f(ax), where a is a positive constant, touches the x-axis at the point (1.5, 0). Find the value of a and the coordinates of the other point where this curve meets the x-axis.
(2)
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Answer
Mark
Notes
Part (a)
positive cubic shape
B1
crosses the x-axis at (−1, 0) and touches it at (3, 0)
B1
meets the y-axis at (0, 9)
B1
Part (b)
the curve from (a) translated 2 units in the positive x-direction
M1
crosses the x-axis at (1, 0) and touches it at (5, 0)
A1
f(−2) = (−1)(−5)2 = −25, so it meets the y-axis at (0, −25)
A1
Part (c)
the touching point moves from x = 3 to x = 3/a = 1.5, so a = 2
B1
the other root moves from −1 to −1/2: (−0.5, 0)
B1
Model sketch
Figure: model sketches of y = f(x) and y = f(x − 2)
Common error: Finding the y-intercept of y = f(x − 2) by moving (0, 9) sideways. The y-intercept is a new point: substitute x = 0 to get f(−2).
A sine wave is shifted sideways. The curve C has equation y = sin(x + 30°), defined for 0 ≤ x ≤ 360°.
(a)
Describe fully the single transformation that maps the graph of y = sin x onto C.
(1)
(b)
Find the coordinates of the points where C meets the coordinate axes.
(3)
(c)
Find the coordinates of the maximum point and of the minimum point of C.
(2)
(d)
Using your answers to (b) and (c), state, with a reason, the number of solutions of the equation sin(x + 30°) = 0.4 for 0 ≤ x ≤ 360°.
(2)
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Answer
Mark
Notes
Part (a)
translation by 30° (30 units) in the negative x-direction
B1
Part (b)
x = 0: sin 30° = 0.5, so (0, 0.5)
B1
sin(x + 30°) = 0 when x + 30° = 180° or 360°
M1
(150°, 0) and (330°, 0)
A1
Part (c)
maximum where x + 30° = 90°: (60°, 1)
B1
minimum where x + 30° = 270°: (240°, −1)
B1
Part (d)
C starts at height 0.5, rises to 1 at 60°, falls to −1 at 240° and rises back to 0.5 at 360°
M1
the line y = 0.4 is crossed once on the way down and once on the way back up: 2 solutions
A1
Model sketch
Figure: model sketch of C with the line y = 0.4
Exam tip: When the variable is shifted, find the new range of the angle first: here 30° ≤ x + 30° ≤ 390°. Every zero, maximum and minimum must have x + 30° in that range.
Figure: model sketch of C₁ (asymptotes x = 3, y = 0) and C₂ (asymptotes x = 0, y = 2)
Common error: Writing the asymptote of y = 6/x + 2 as x = 2. Adding 2 outside the fraction moves the curve up, so it is the horizontal asymptote that moves: y = 2.
A cubic curve C has equation y = f(x). C crosses the x-axis at (−3, 0), touches the x-axis at (2, 0) and meets the y-axis at (0, 24). There are no other points where C meets the coordinate axes.
Figure: sketch of the curve C
(a)
Find f(x) in factorised form.
(3)
(b)
Show that f(x) = 2x3 − 2x2 − 16x + 24.
(2)
(c)
Write down the coordinates of the points where the curve y = f(−x) meets the coordinate axes.
(2)
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Answer
Mark
Notes
Part (a)
f(x) = k(x + 3)(x − 2)2
M1
x = 0: k × 3 × 4 = 24
M1
k = 2, so f(x) = 2(x + 3)(x − 2)2
A1
Part (b)
(x + 3)(x2 − 4x + 4) = x3 − x2 − 8x + 12
M1
multiplying by 2 gives 2x3 − 2x2 − 16x + 24
A1
Part (c)
crosses the x-axis at (3, 0) and touches it at (−2, 0)
B1
meets the y-axis at (0, 24)
B1
Exam tip: A touching point on the x-axis means a squared factor. Write the general form k(x − p)(x − q)2 first, then use the third point to find k.
Figure 1 shows the curve with equation y = f(x), where f(x) is a cubic polynomial, and the line with equation y = x − 1. The curve meets the x-axis only at the two points marked and passes through the point P(2, −6).
(a)
Explain why f(x) = a(x − 1)2(x + 4), and find the value of the constant a.
(3)
(b)
The curve y = f(x − k), where k is a constant, passes through the origin. Find the possible values of k.
(2)
(c)
Find the exact coordinates of all the points where the line y = x − 1 meets the curve y = f(x).
(4)
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Answer
Mark
Notes
Part (a)
the curve touches the x-axis at x = 1, so (x − 1) is a repeated factor, and it crosses at x = −4, so (x + 4) is a factor
Common error: Writing (x + 1)2 for the repeated factor. The curve touches the axis at x = 1, so the factor is (x − 1)2: the sign in the bracket is opposite to the root.
Figure 1 shows the curve C with equation y = k/(x − p) + q, where k, p and q are constants and k > 0. The asymptotes of C are shown dashed. C passes through the point A(4, 1). Figure 1 also shows the line L with equation y = 1 − x.
(a)
Using Figure 1, write down the values of p and q.
(1)
(b)
Describe fully the translation that maps the curve y = k/x onto C, and find the value of k.
(3)
(c)
Find the coordinates of the points where C meets the coordinate axes.
(2)
(d)
Show algebraically that L does not meet C.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
asymptotes x = 2 and y = −1: p = 2, q = −1
B1
Part (b)
translation by 2 units in the positive x-direction and 1 unit in the negative y-direction, vector (2, −1)
B1
at A: k/(4 − 2) − 1 = 1
M1
k = 4
A1
Part (c)
x = 0: y = 4/(−2) − 1 = −3, so (0, −3)
B1
y = 0: 4/(x − 2) = 1, so (6, 0)
B1
Part (d)
4/(x − 2) − 1 = 1 − x, so 4 = (2 − x)(x − 2)
M1
(x − 2)2 = −4
A1
a square cannot be negative, so there are no real solutions and L does not meet C
A1
Exam tip: A translation moves the asymptotes with the curve. y = k/x has asymptotes x = 0 and y = 0, so the dashed lines in the figure give the translation straight away.
In Figure 1, the left-hand diagram shows the curve y = f(x), where f(x) = a(x + 3)(x − 1)2 and a is a constant. The right-hand diagram shows the image of this curve under a single transformation. The labelled points are where each curve meets the coordinate axes.
(a)
Describe fully the single transformation, and write down the equation of the transformed curve in terms of f.
(2)
(b)
Find the value of a.
(2)
(c)
The curve y = f(x + c), where c is a constant, touches the x-axis at the origin. Find the value of c and the coordinates of the other point where this curve meets the x-axis.
(2)
(d)
Show that the transformed curve has equation y = 4x3 + 2x2 − 5x + 3/2.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
stretch parallel to the x-axis, scale factor ½
B1
y = f(2x)
B1
Part (b)
f(0) = a(3)(−1)2 = 3a = 1.5
M1
a = ½
A1
Part (c)
the touching point (1, 0) moves to (0, 0), a translation of 1 unit in the negative x-direction: c = 1
Common error: Calling this a stretch of scale factor 2. Every x-coordinate has been halved while the y-intercept stays at 1.5, so the scale factor parallel to the x-axis is ½ and the equation is y = f(2x).
Figure 1 shows the graph of y = f(x), −4 ≤ x ≤ 4. The graph consists of three straight line segments joining the points A(−4, 0), B(−2, 4), C(2, −2) and D(4, 0).
(a)
Find the value of f(0) and the x-coordinate of the point between B and C where the graph crosses the x-axis.
(2)
(b)
The lowest point of the graph of y = f(x) + c lies on the x-axis. Write down the value of the constant c, and state the number of points where the graph of y = f(x) + c meets the x-axis.
(2)
(c)
The graph of y = f(px), where p is a positive constant, has an endpoint at (−8, 0). Find p and write down the coordinates of the images of B and C.
(2)
(d)
The highest point of the graph of y = qf(x), where q is a constant, is (2, 3). Find q and write down the coordinates of the lowest point of this graph.
(2)
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Answer
Mark
Notes
Part (a)
gradient of BC = −6/4 = −1.5, so f(0) = 4 − 1.5 × 2 = 1
B1
4 − 1.5(x + 2) = 0 gives x = 2/3
B1
Part (b)
c = 2
B1
1 point, the image of C, (2, 0), since the images of A and D are at height 2
B1
Part (c)
A(−4, 0) maps to (−4/p, 0) = (−8, 0), so p = ½
B1
(−4, 4) and (4, −2)
B1
Part (d)
the highest point is the image of C: −2q = 3, so q = −1.5
B1
lowest point is the image of B: (−2, −6)
B1
Common error: Multiplying the x-coordinates by p for y = f(px). A stretch inside the bracket divides each x-coordinate by p, so with p = ½ they double.
Figure 1 shows four curves. Curve A has equation y = k/x + c. Curve B has equation y = k/(x + p)2. Curve C has equation y = a(x − p)2(x − q). Curve D has equation y = cos(x + b), 0 ≤ x ≤ 360°, where 0 < b < 90°. The constants may take different values in different curves. Use the labelled features in Figure 1 to answer each part.
(a)
For curve A, find the values of c and k.
(2)
(b)
For curve B, find the values of p and k.
(2)
(c)
For curve C, find the values of a, p and q.
(2)
(d)
For curve D, find the value of b and the exact y-coordinate of the point where curve D meets the y-axis.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
horizontal asymptote y = 2: c = 2
B1
through (−1, 0): 0 = −k + 2, k = 2
B1
Part (b)
vertical asymptote x = −1, so x + p = 0 when x = −1: p = 1
B1
through (0, 3): k/1 = 3, k = 3
B1
Part (c)
touches at −2 and crosses at 1: p = −2, q = 1
B1
through (0, 4): a(4)(−1) = 4, a = −1
B1
Part (d)
the first zero of cos is where x + b = 90°, and x = 60°: b = 30°
B1
cos 30° = √3/2
B1
Common error: Reading the asymptote x = −1 on curve B and writing (x − 1)2. The bracket must be zero at the asymptote, so it is (x + 1)2 and p = 1.
The function g is defined by g(x) = 4x − x3. Its graph has rotational symmetry about the origin.
(a)
Factorise g(x) completely.
(2)
(b)
Sketch the curve y = g(x), showing the coordinates of the points where it meets the coordinate axes.
(3)
(c)
On the same axes, sketch the curve y = 3/x.
(2)
(d)
Show that the x-coordinates of the points where the two curves meet satisfy x4 − 4x2 + 3 = 0.
(2)
(e)
Hence find the exact coordinates of all the points where the two curves meet.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
x(4 − x2)
M1
x(2 − x)(2 + x)
A1
Part (b)
negative cubic shape: from top left to bottom right
B1
crosses the x-axis at (−2, 0), (0, 0) and (2, 0)
B1
minimum between −2 and 0 and maximum between 0 and 2, passing through the origin
B1
Part (c)
two branches in the first and third quadrants
B1
asymptotes x = 0 and y = 0, no axis intercepts
B1
Part (d)
4x − x3 = 3/x, so 4x2 − x4 = 3 (x ≠ 0)
M1
x4 − 4x2 + 3 = 0
A1
Part (e)
(x2 − 1)(x2 − 3) = 0
M1
x = ±1 or x = ±√3
A1
using y = 3/x: (1, 3) and (−1, −3)
A1
(√3, √3) and (−√3, −√3)
A1
Model sketch
Figure: model sketch of y = g(x) and y = 3/x with their four meeting points
Exam tip: Use the sketch to predict the answer before the algebra: the hump between 0 and 2 can cut the branch of 3/x twice, and by symmetry the same happens between −2 and 0 — four points in all.
f(x) = x3 − 12x The curve y = f(x) has a maximum point at (−2, 16) and a minimum point at (2, −16).
(a)
Find the exact coordinates of the points where the curve y = f(x) meets the x-axis.
(2)
(b)
Sketch the curve y = f(x), labelling the turning points and the points where it meets the x-axis.
(2)
(c)
The curve y = f(x) + k, where k is a constant, touches the x-axis. Find the two possible values of k and, for each value, the coordinates of the point where the curve touches the x-axis.
(4)
(d)
Write down the coordinates of the turning points of the curve y = f(½x), and the exact coordinates of the points where it meets the x-axis.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x(x2 − 12) = 0
M1
(0, 0), (2√3, 0) and (−2√3, 0)
A1
Part (b)
positive cubic through the origin with maximum (−2, 16) and minimum (2, −16) labelled
B1
x-intercepts −2√3, 0 and 2√3 labelled
B1
Part (c)
the curve touches the x-axis only where a turning point lies on the axis, so the translation must move a turning point onto y = 0
M1
k = 16, which moves the minimum to the axis
A1
it touches at (2, 0)
A1
k = −16, touching at (−2, 0)
A1
Part (d)
stretch parallel to the x-axis, scale factor 2, so x-coordinates are doubled
M1
maximum (−4, 16), minimum (4, −16)
A1
(0, 0), (4√3, 0) and (−4√3, 0)
A1
Model sketch
Figure: model sketch of y = f(x)
Common error: For f(½x), halving the x-coordinates. The new curve reaches f(p) when ½x = p, that is at x = 2p, so the x-coordinates double.
for x > −2 the line rises from 2 while C falls from ∞ to 0, so they meet once
M1
for −4 < x < −2 the line is below 2 while C is above 9/4, and for x ≤ −4 the line is not positive while C is, so no more meetings: 1 real root
A1
Model sketch
Figure: model sketch of C (asymptotes x = −2 and y = 0) with the line y = x + 4
Exam tip: When the reciprocal graph has a shifted asymptote, split the counting argument at the asymptote and compare the two graphs on each side separately.
The cubic function f is given by f(x) = x3 + 2x2 − 15x. Its graph is used to study a translation, a stretch and an equation of degree four.
(a)
Factorise f(x) completely.
(2)
(b)
Sketch the curve y = f(x), showing the coordinates of the points where it meets the coordinate axes.
(3)
(c)
The curve y = f(x + k), where k is a non-zero constant, passes through the origin. Find the possible values of k.
(2)
(d)
The curve y = f(ax), where a is a positive constant, crosses the positive x-axis at (12, 0). Find a and the coordinates of the other points where this curve meets the x-axis.
(3)
(e)
On a copy of your sketch from (b), sketch the curve y = −2/x. Hence state the number of real solutions of the equation x4 + 2x3 − 15x2 + 2 = 0, giving a reason.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x(x2 + 2x − 15)
M1
x(x + 5)(x − 3)
A1
Part (b)
positive cubic shape
B1
crosses the x-axis at (−5, 0), (0, 0) and (3, 0)
B1
maximum between −5 and 0, minimum between 0 and 3, passing through the origin
B1
Part (c)
f(0 + k) = 0, so f(k) = 0
M1
k = −5 or k = 3
A1
Part (d)
the positive root moves from 3 to 3/a, so 3/a = 12
M1
a = ¼
A1
(0, 0) and (−20, 0)
A1
Part (e)
y = −2/x: branches in the second and fourth quadrants, asymptotic to the axes
B1
multiplying f(x) = −2/x by x (x ≠ 0) gives x4 + 2x3 − 15x2 + 2 = 0, so the solutions are the meeting points
M1
the curves meet twice for −5 < x < 0 and twice for 0 < x < 3: 4 real solutions
A1
Model sketch
Figure: model sketch of y = f(x) and y = −2/x (they meet four times)
Common error: Thinking f(x + k) passes through the origin when k is a root with the sign changed. Substitute x = 0: the condition is f(k) = 0, so k is itself a root.
Figure 1 shows the curve y = f(x), where f(x) is a cubic polynomial, and the curve y = 8/x2. The cubic crosses the x-axis at the three points marked and meets the y-axis at (0, 6). The two curves meet at P(−1, 8) and at other points.
(a)
Using Figure 1, find f(x), giving your answer in the form x3 + bx2 + cx + d.
(3)
(b)
Show that the curve y = f(x − 2) passes through the origin, and write down the other points where it meets the x-axis.
(2)
(c)
Show that the x-coordinates of the points where the two curves meet satisfy x5 − 2x4 − 5x3 + 6x2 − 8 = 0.
(2)
(d)
Using Figure 1, explain why this equation has exactly three real roots.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
f(x) = a(x + 2)(x − 1)(x − 3)
B1
f(0) = a(2)(−1)(−3) = 6a = 6, so a = 1
M1
f(x) = x3 − 2x2 − 5x + 6
A1
Part (b)
at x = 0: f(−2) = 0, so the curve passes through (0, 0)
B1
the roots 1 and 3 move 2 to the right: (3, 0) and (5, 0)
B1
Part (c)
x3 − 2x2 − 5x + 6 = 8/x2, x ≠ 0
M1
multiplying by x2: x5 − 2x4 − 5x3 + 6x2 − 8 = 0
A1
Part (d)
for −2 < x < 0 the curves meet twice: at P and at a point with −2 < x < −1
B1
for x > 3 the cubic rises from 0 while 8/x2 falls, so they meet once
B1
no other meetings: for x < −2 and 1 < x < 3 the cubic is negative but 8/x2 > 0, and for 0 < x ≤ 1 the cubic is at most 6 while 8/x2 ≥ 8; so 3 real roots
A1
Exam tip: When a count of roots must be explained, split the x-axis into intervals using the roots and asymptotes, and say on which side of the other curve each one lies in every interval.
Figure 1 shows the curve C with equation y = k/(x − p)2 + q, where k, p and q are constants. The asymptotes of C are shown dashed, and C passes through the point (0, 1).
(a)
Using Figure 1, find the values of p, q and k.
(3)
(b)
Describe fully the single transformation that maps the curve y = k/x2 onto C.
(1)
(c)
Find the exact x-coordinates of the points where C meets the x-axis.
(3)
(d)
Find the set of values of the constant m for which the equation k/(x − p)2 + q = m has no real solutions.
(2)
(e)
Describe fully the single transformation that maps C onto the curve with equation y = 3/(x + 1)2.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
vertical asymptote x = 1: p = 1
B1
horizontal asymptote y = −2: q = −2
B1
through (0, 1): k/1 − 2 = 1, so k = 3
B1
Part (b)
translation by 1 unit in the positive x-direction and 2 units in the negative y-direction, vector (1, −2)
B1
Part (c)
3/(x − 1)2 = 2
M1
(x − 1)2 = 3/2
A1
x = 1 ± √6/2
A1
Part (d)
C takes every value greater than −2 but never reaches its asymptote y = −2
M1
m ≤ −2
A1
Part (e)
translation
B1
by 2 units in the negative x-direction and 2 units in the positive y-direction, vector (−2, 2)
B1
Common error: Leaving m = −2 out of the answer to (d). The curve never reaches its horizontal asymptote, so the equation has no solutions when m = −2 as well.
Figure 1 shows the curve y = f(x), where f(x) is a cubic polynomial, and the line L with equation y = 8 + 2x. The curve meets the x-axis at the three points marked and meets the y-axis at (0, 8). L meets the curve at (0, 8) and at (5, 18).
(a)
Using Figure 1, find f(x) in factorised form, and show that f(x) = x3 − 5x2 + 2x + 8.
(4)
(b)
Show that the x-coordinates of the points where L meets the curve satisfy x2(x − 5) = 0, and explain what this tells you about L and the curve at the point (0, 8).
(3)
(c)
Find the set of values of the constant m for which the line y = mx + 8 meets the curve at three distinct points.
(4)
(d)
Write down the coordinates of the points where the curve y = f(½x) meets the coordinate axes.
the quadratic needs two distinct real roots: 25 − 4(2 − m) > 0
M1
m > −17/4
A1
neither root may be x = 0, so 2 − m ≠ 0: m > −17/4, m ≠ 2
A1
Part (d)
x-axis: (−2, 0), (4, 0), (8, 0)
B1
y-axis: (0, 8)
B1
Exam tip: A repeated root of the equation “curve = line” means the line touches the curve there. Here x2 is a factor, so L is a tangent at x = 0, which matches the figure.
Figure 1 shows the curve y = f(x), where f(x) = (x + 1)2(x − 2), and the curve y = g(x), shown dashed, where g(x) = af(x + b) for constants a and b. The curve y = f(x) has a maximum point at (−1, 0) and a minimum point at (1, −4). The points labelled on y = g(x) are its minimum point (1, 0), its maximum point (3, 2) and the point (4, 0).
(a)
Find the values of a and b.
(3)
(b)
Describe fully a sequence of transformations that maps the curve y = f(x) onto the curve y = g(x).
(2)
(c)
Show that g(x) = −½x3 + 3x2 − (9/2)x + 2, and write down the coordinates of the point where y = g(x) meets the y-axis.
(3)
(d)
Show that the x-coordinates of the points where the two curves meet satisfy 3x(x − 1)2 = 8.
(2)
(e)
Using Figure 1, state the number of points where the two curves meet.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
every point moves 2 units to the right, so x + b = x − 2: b = −2
B1
the minimum (1, −4) of f maps to the maximum (3, 2) of g, so −4a = 2
M1
a = −½
A1
Part (b)
translation by 2 units in the positive x-direction
B1
stretch parallel to the y-axis, scale factor ½, and reflection in the x-axis (allow stretch parallel to the y-axis, scale factor −½)
×2 and rearrange: 3x3 − 6x2 + 3x = 8, so 3x(x − 1)2 = 8
A1
Part (e)
1
B1
Common error: Taking b = 2 because the curve has moved 2 units to the right. For g(x) = af(x + b), a shift in the positive x-direction needs x + b = x − 2, so b = −2.
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