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Sketching polynomial graphs from their factorised form and the reciprocal graphs \(y = \dfrac{a}{x}\) and \(y = \dfrac{a}{x^2}\) with their asymptotes. You use points of intersection to count or find solutions of equations, and work with proportional relationships and their graphs.
Squared factor: the curve touches the \(x\)-axis there
\(y = \dfrac{a}{x}\) and \(y = \dfrac{a}{x^2}\): asymptotes \(x = 0\) and \(y = 0\)
\(y = \dfrac{a}{x^2}\),\(a \gt 0\): always above the \(x\)-axis
Number of real solutions of \(f(x) = g(x)\) = number of intersections
\(y \propto x^n \Leftrightarrow y = kx^n\)
Notes
Polynomial graphs
diagram
Roots come from the factorised form (put \(y = 0\)); the \(y\)-intercept comes from putting \(x = 0\)
Shape comes from the leading term: a positive cubic goes from bottom left to top right, a negative cubic from top left to bottom right. A quartic with positive \(x^4\) coefficient has both ends going up.
A repeated root \((x - a)^2\) touches the \(x\)-axis at \(x = a\) (a turning point). A triple root \((x - a)^3\) flattens out and crosses.
Example: \(y = (x + 1)(x - 2)^2\) crosses at \(x = -1\), touches at \(x = 2\) and meets the \(y\)-axis at \((0, 4)\)
A single factor gives a crossing point; a squared factor gives a point where the curve touches the axis.
Reciprocal graphs
diagram
\(y = \dfrac{a}{x}\) with \(a \gt 0\): branches in the first and third quadrants. With \(a \lt 0\): second and fourth quadrants.
\(y = \dfrac{a}{x^2}\) with \(a \gt 0\): both branches above the \(x\)-axis, symmetric about the \(y\)-axis.
Both curves have the axes x = 0 and y = 0 as asymptotes. a/x² is never negative.
Both have asymptotes \(x = 0\) and \(y = 0\): lines the curve gets closer and closer to but never reaches. State them as equations.
Proportion: \(y = kx\) is a straight line through the origin; \(y = \dfrac{k}{x}\) is inverse proportion; \(y \propto x^n\) means \(y = kx^n\)
Intersections
The solutions of \(f(x) = g(x)\) are the \(x\)-coordinates of the points where \(y = f(x)\) and \(y = g(x)\) meet.
So you can state the number of real solutions of an equation by counting intersections on a sketch.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Sketch a polynomial, showing where it meets the axes
3 marksGrade D
Find the roots from the factors and note any repeated root.
Find the \(y\)-intercept.
Use the leading term for the shape and label all intercepts with coordinates.
Example. Sketch the curve with equation \(y = x(x - 3)^2\), showing the coordinates of the points where it meets the coordinate axes.
Show the model answerHide the model answer
Cubic shape rising from bottom left to top right
(B1)
Passes through \((0, 0)\)
(B1)
Touches the \(x\)-axis at \((3, 0)\), a minimum point
(B1)
Use a sketch to find the number of solutions
4 marksGrade C
Sketch both graphs on the same axes, with key points and asymptotes.
Count the points of intersection.
State the number of solutions with a reason.
Example. (a) On the same axes, sketch the curves \(y = x^2(x - 3)\) and \(y = \dfrac{2}{x}\). (b) Hence state the number of real solutions of the equation \(x^3(x - 3) = 2\)
Show the model answerHide the model answer
(a) Cubic touching the origin and crossing the \(x\)-axis at \((3, 0)\)
(B1)
Reciprocal curve in the first and third quadrants, with the axes as asymptotes
(B1)
(b) \(x^3(x - 3) = 2\) is equivalent to \(x^2(x - 3) = \dfrac{2}{x}\) (\(x \ne 0\)), so count intersections
(M1)
The curves meet once for \(x \gt 3\) and once for \(x \lt 0\), so there are 2 real solutions.
(A1)
Don’t lose marks
Drawing a squared factor as a crossing point instead of a touching point.
Letting a reciprocal curve cross, touch or curl away from its asymptote.
Drawing \(y = \dfrac{a}{x^2}\) (with \(a \gt 0\)) with a branch below the \(x\)-axis.
Not giving coordinates such as \((3, 0)\) and \((0, 0)\) when the question asks for them.
More tips
Memory tricks
Sketch checklist: shape, \(x\)-intercepts, \(y\)-intercept, asymptotes labelled with equations.
For \(y = (x - a)(x - b)(x - c)\), the \(y\)-intercept is \((-a)(-b)(-c)\)
Exam technique
'Sketch' means show the key features (shape, intercepts, asymptotes), not plot points accurately; label everything the question asks for.
After 'hence' with a sketch, count the intersections of the graphs you drew and give a reason.
Asymptotes must be stated as equations, e.g. \(x = 0\), not just drawn.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Sketch a cubic from its factorised formMark the roots and the \(y\)-intercept; a positive \(x^3\) coefficient goes from bottom left to top right.
Grade D
Show repeated roots correctly on a sketchA squared factor such as \((x - 2)^2\) means the curve touches the \(x\)-axis at \(x = 2\) without crossing.
Grade D
Sketch reciprocal graphs and their asymptotes\(y = \dfrac{a}{x}\) and \(y = \dfrac{a}{x^2}\) both have asymptotes \(x = 0\) and \(y = 0\)
Grade D
Use proportional relationships and their graphs\(y \propto x^n\) means \(y = kx^n\); find \(k\) from one pair of values.
Grade C
Use intersections to count solutionsThe number of points where \(y = f(x)\) and \(y = g(x)\) meet is the number of real solutions of \(f(x) = g(x)\)
Grade A
Find values of k for a number of intersectionsUse a sketch (and sometimes the discriminant) to find where a line or curve meets a graph a given number of times.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve \(C\) has equation \(y = (x + 2)(x - 4)^2\).
(a) Sketch \(C\), showing clearly the coordinates of all the points where \(C\) meets the coordinate axes.[3]
(b) Using your sketch, find the set of values of \(x\) for which \(y \gt 0\).[2]
Show the answer and mark scheme
(a)Answer: A cubic with a positive leading coefficient (rising from bottom left to top right) which crosses the \(x\)-axis at \((-2, 0)\), touches the \(x\)-axis at \((4, 0)\), and meets the \(y\)-axis at \((0, 32)\).
B1 for the correct shape: a cubic with a positive leading coefficient (rising from bottom left to top right)
B1 for the correct behaviour at the \(x\)-axis: it crosses the \(x\)-axis at \((-2, 0)\); touches the \(x\)-axis at \((4, 0)\)
B1 for the \(y\)-intercept \((0, 32)\)
Worked solution: \(y = (x + 2)(x - 4)^2\). Expanding would give a leading term \(x^3\), so the curve is a cubic with a positive leading coefficient. Roots: crosses the \(x\)-axis at \((-2, 0)\); touches the \(x\)-axis at \((4, 0)\). When \(x = 0\), \(y = 32\), so the curve meets the \(y\)-axis at \((0, 32)\).
(b)Answer: \(-2 \lt x \lt 4\) or \(x \gt 4\)
M1 for using the \(x\)-intercepts as critical values and choosing regions consistent with their sketch
A1 for \(-2 \lt x \lt 4\) or \(x \gt 4\)
Worked solution: From the sketch, the curve is above the \(x\)-axis when \(-2 \lt x \lt 4\) or \(x \gt 4\).
Question 2Medium6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve \(C\) has equation \(y = (x + 4)(x - 1)^2(x - 2)\).
(a) Sketch \(C\), showing clearly the coordinates of all the points where \(C\) meets the coordinate axes.[4]
(b) Using your sketch, find the set of values of \(x\) for which \(y \ge 0\).[2]
Show the answer and mark scheme
(a)Answer: A quartic with a positive leading coefficient (with both ends going up) which crosses the \(x\)-axis at \((-4, 0)\) and \((2, 0)\), touches the \(x\)-axis at \((1, 0)\), and meets the \(y\)-axis at \((0, -8)\).
B1 for the correct shape: a quartic with a positive leading coefficient (with both ends going up)
B1 for the correct behaviour at the \(x\)-axis: it crosses the \(x\)-axis at \((-4, 0)\) and \((2, 0)\); touches the \(x\)-axis at \((1, 0)\)
B1 for all \(x\)-intercepts labelled correctly (as coordinates or values)
B1 for the \(y\)-intercept \((0, -8)\)
Worked solution: \(y = (x + 4)(x - 1)^2(x - 2)\). Expanding would give a leading term \(x^4\), so the curve is a quartic with a positive leading coefficient. Roots: crosses the \(x\)-axis at \((-4, 0)\) and \((2, 0)\); touches the \(x\)-axis at \((1, 0)\). When \(x = 0\), \(y = -8\), so the curve meets the \(y\)-axis at \((0, -8)\).
(b)Answer: \(x \le -4\) or \(x = 1\) or \(x \ge 2\)
M1 for using the \(x\)-intercepts as critical values and choosing regions consistent with their sketch
A1 for \(x \le -4\) or \(x = 1\) or \(x \ge 2\)
Worked solution: From the sketch, the curve is above the \(x\)-axis (or on it) when \(x \le -4\) or \(x = 1\) or \(x \ge 2\).
Question 3Hard9 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The diagram shows a sketch of the curve \(C\) with equation \(y = x^2(x - 5)\). The line \(l\) has equation \(y = kx\), where \(k\) is a constant.
[object Object]
(a) Show that the \(x\)-coordinates of the points where \(l\) meets \(C\) satisfy \(x = 0\) or \(x^2 - 5x - k = 0\).[2]
(b) Find the set of values of \(k\) for which \(l\) meets \(C\) at exactly three distinct points.[4]
(c) Find the values of \(k\) for which \(l\) meets \(C\) at exactly two distinct points. For each value, state the coordinates of the points of intersection.[3]
M1 for equating and taking out the common factor \(x\): \(x\left(x^2 - 5x - k\right) = 0\)
A1* for the given result (not dividing by \(x\) without considering \(x = 0\))
Worked solution: \(x^2(x - 5) = kx\) so \(x^{3} - 5x^{2} - kx = x\left(x^2 - 5x - k\right) = 0\). Hence \(x = 0\) or \(x^2 - 5x - k = 0\).
(b)Answer: \(-\frac{25}{4} \lt k \lt 0\) or \(k \gt 0\)
M1 for requiring \(x^2 - 5x - k = 0\) to have two distinct real roots: \(25 + 4k \gt 0\)
A1 for \(k \gt -\frac{25}{4}\)
M1 for recognising that \(x = 0\) is a root of \(x^2 - 5x - k = 0\) when \(k = 0\), so \(k \ne 0\)
A1 for \(-\frac{25}{4} \lt k \lt 0\) or \(k \gt 0\)
Worked solution: The point \(x = 0\) always lies on both graphs. For two further distinct points, \(x^2 - 5x - k = 0\) needs two distinct real roots, neither of them 0. Discriminant: \((-5)^2 + 4k = 25 + 4k \gt 0\), so \(k \gt -\frac{25}{4}\). A root is 0 exactly when \(k = 0\), so \(k \ne 0\). So \(-\frac{25}{4} \lt k \lt 0\) or \(k \gt 0\).
(c)Answer: \(k = 0\): \((0, 0)\) and \((5, 0)\); \(k = -\frac{25}{4}\): \((0, 0)\) and \(\left(\frac{5}{2}, -\frac{125}{8}\right)\)
B1 for \(k = 0\) with the points \((0, 0)\) and \((5, 0)\) (the quadratic becomes \(x^{2} - 5x = 0\))
M1 for \(k = -\frac{25}{4}\), from \(25 + 4k = 0\) (a repeated root)
A1 for the points \((0, 0)\) and \(\left(\frac{5}{2}, -\frac{125}{8}\right)\)
Worked solution: Exactly two distinct points occur when \(x^2 - 5x - k = 0\) has 0 as a root (\(k = 0\), giving \(x = 0\) or \(x = 5\)) or has a repeated root (\(k = -\frac{25}{4}\), giving \(x = \frac{5}{2}\)). \(k = 0\): the points are \((0, 0)\) and \((5, 0)\) (the line is the \(x\)-axis, which touches \(C\) at \(O\)). \(k = -\frac{25}{4}\): the points are \((0, 0)\) and \(\left(\frac{5}{2}, -\frac{125}{8}\right)\) (the line \(y = -\frac{25}{4}x\) is a tangent to \(C\) at \(\left(\frac{5}{2}, -\frac{125}{8}\right)\)).