Polynomial and rational graphs: notes and questions
Revision for sketching cubics given as a product of linear factors, the modulus of a cubic, cubic inequalities from a sketch, and (4PM1) rational functions with asymptotes and solving equations with graphs. Both boards sketch cubics from their factors. Graphs of y = |f(x)| and cubic inequalities are Cambridge 0606 only. Rational functions with asymptotes and solving equations with graphs are Edexcel 4PM1 only.
For y = a(x − p)(x − q)(x − r), the curve crosses the x-axis at x = p, q and r. Put x = 0 to find where it meets the y-axis.
If a > 0 the cubic rises to the right (y → ∞ as x → ∞); if a < 0 it falls to the right.
A squared factor (x − p)² means the curve touches the x-axis at x = p instead of crossing it.
To solve f(x) > 0 for a cubic, sketch it and read off where the curve is above the x-axis.
For y = |f(x)|, reflect any part of the graph below the x-axis in the x-axis; points on the axis stay where they are.
For y = (ax + b)/(cx + d) (4PM1): the vertical asymptote is where cx + d = 0, and the horizontal asymptote is y = a/c.
To solve an equation with a graph (4PM1), rearrange it so that one side is the curve already drawn; the other side is the line to draw. The roots are the x-coordinates of the points where they meet.
Watch out: Label every intercept on a sketch with its coordinates, including the y-intercept. Marks are given for the labels as well as the shape.
Worked example· 3 marks
Sketch the curve \(y = (x + 2)(x - 1)(x - 3)\), stating the coordinates of the points where it meets the axes.
B1 positive cubic shape: falls to the left, rises to the right
B1 (−2, 0), (1, 0) and (3, 0)
B1 (0, 6)
y = 0 when x + 2 = 0, x − 1 = 0 or x − 3 = 0: the curve crosses the x-axis at x = −2, 1 and 3.
x = 0 gives y = (2)(−1)(−3) = 6, so it meets the y-axis at (0, 6).
Multiplying out, the x³ term is +x³, so the curve comes up from the bottom left and rises to the top right.
Sketch: up through (−2, 0), over a maximum, down through (0, 6) and (1, 0), a minimum, then up through (3, 0).
Answer: meets the \(x\)-axis at \((-2, 0)\), \((1, 0)\), \((3, 0)\) and the \(y\)-axis at \((0, 6)\)
Polynomial and rational graphs questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 4 marks06064PM1no calculator
The curve \(C\) has equation \(y = (2x - 1)(x + 3)(4 - x)\).
(a) Find the coordinates of the points where \(C\) meets the axes.
(b) Describe what happens to \(y\) as \(x\) becomes large and positive.
Show the mark scheme and worked solution
B1 (a) (½, 0), (−3, 0) and (4, 0)
B1 (a) (0, −12)
M1 (b) leading term −2x³
A1 (b) y → −∞
Roots: 2x − 1 = 0 gives x = ½; x + 3 = 0 gives x = −3; 4 − x = 0 gives x = 4.
x = 0: y = (−1)(3)(4) = −12, so C meets the y-axis at (0, −12).
The x³ term is (2x)(x)(−x) = −2x³. With a negative coefficient, y → −∞ as x → ∞: the curve falls to the right.
Answer: (a) \((\frac{1}{2}, 0)\), \((-3, 0)\), \((4, 0)\) and \((0, -12)\) (b) \(y\) becomes large and negative
Question 2 · 4 marks4PM1no calculator
The curve \(C\) has equation \(y = \frac{2x + 3}{x - 1}\).
(a) Write down the equations of the asymptotes of \(C\).
(b) Find the coordinates of the points where \(C\) meets the axes.
Show the mark scheme and worked solution
B1 (a) x = 1
B1 (a) y = 2
B1 (b) (−3/2, 0)
B1 (b) (0, −3)
The denominator is zero at x = 1, so x = 1 is a vertical asymptote.
For large x, y is close to 2x/x, which is 2, so y = 2 is the horizontal asymptote.
y = 0 when 2x + 3 = 0, so x = −3/2: the point (−3/2, 0).
x = 0 gives y = 3/(−1), which is −3: the point (0, −3).
Answer: (a) \(x = 1\) and \(y = 2\) (b) \((-\frac{3}{2}, 0)\) and \((0, -3)\)
Question 3 · 4 marks06064PM1no calculator
The curve \(y = a(x + 1)(x - 2)(x - 5)\) passes through the point \((0, 20)\).
(a) Find the value of \(a\).
(b) Write the equation of the curve in the form \(y = px^3 + qx^2 + rx + s\).
Show the mark scheme and worked solution
M1 (a) substitute (0, 20)
A1 (a) a = 2
M1 (b) expand the three brackets
A1 (b) y = 2x³ − 12x² + 6x + 20
At x = 0: a(1)(−2)(−5) = 10a, and this equals 20, so a = 2.
Sketch the curve \(y = (x - 2)^2(x + 1)\), showing the coordinates of the points where it meets the axes and how it behaves at each of them.
Show the mark scheme and worked solution
B1 crosses the x-axis at (−1, 0)
B1 touches the x-axis at (2, 0)
B1 meets the y-axis at (0, 4)
B1 positive cubic shape
The factor (x + 1) appears once, so the curve crosses the x-axis at x = −1.
The factor (x − 2) is squared, so the curve touches the x-axis at x = 2 (a minimum there).
x = 0 gives y = (−2)²(1), which is 4: the point (0, 4).
The x³ coefficient is +1, so the curve rises from the bottom left, crosses at −1, passes through (0, 4) and a maximum, comes down to touch at (2, 0), then rises.
Answer: crosses at \((-1, 0)\), touches at \((2, 0)\), meets the \(y\)-axis at \((0, 4)\)
9 more polynomial and rational graphs questions
Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.
IGCSE Math Revision