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A-Level Graphs and their transformations
What the A-Level syllabus expects for Graphs and their transformations, and how to practise it.
What the syllabus expects
- Sketching a supplied function using a graphing calculator or graphing software
- Key features (symmetry, axis intercepts, turning points, asymptotes) of standard curves including y² = ax, the ellipse x²/a² + y²/b² = 1, the hyperbolas x²/a² − y²/b² = 1 and y²/a² − x²/b² = 1, and the rational curves y = (ax+b)/(cx+d) and y = (ax²+bx+c)/(dx+e)
- Working out asymptote equations, axes of symmetry, and the permitted ranges of x and/or y
- How the graph of y = f(x) changes under y = af(x), y = f(x)+a, y = f(x+a) and y = f(ax), and under combinations of these
- Linking the graphs of y = |f(x)|, y = f(|x|) and y = 1/f(x) back to the graph of y = f(x)
- Basic parametric equations and the curves they produce
How it's examined
Questions on this topic most often ask you to sketch, find, compare, prove. About 7% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
The curve C is given by y = ln(x − 2) + 3. Draw a sketch of C, giving the exact coordinates of any point(s) where it meets the axes and the equation of its asymptote.
Show the worked answer
y = ln(x - 2) + 3. Domain: x - 2 > 0, i.e. x > 2. Vertical asymptote: x = 2 (as x -> 2+, y -> -infinity). No y-axis intercept (x = 0 is outside the domain x > 2). x-axis intercept: set y = 0 => ln(x - 2) = -3 => x - 2 = e⁻³ => x = 2 + e⁻³, giving the point (2 + e⁻³, 0). Shape: an increasing logarithmic curve, rising steeply just to the right of x = 2 and flattening as x increases, crossing the x-axis at (2 + e⁻³, 0).
Example 2 (2 marks)
On one diagram, sketch y = ln x and y = x - 5, giving any asymptote equations and the x-coordinates where the two graphs cross.
Show the worked answer
y = ln x: increasing curve through (1,0), defined only for x>0, with the y-axis as a vertical asymptote (x = 0), rising slowly and passing (e,1). y = x - 5: straight line, gradient 1, crossing the x-axis at (5,0) and the y-axis at (0,-5). The two graphs meet where ln x = x - 5. This has two solutions: one for small x where ln x rises steeply from -infinity to overtake the line, and one for larger x where the line finally overtakes the log curve. Solving numerically gives x ~ 0.0067 and x ~ 6.95. Mark these two intersection points on the sketch.
Example 3 (3 marks)
Sketch curve C, clearly showing its asymptotes and its stationary points.
Show the worked answer
Curve C is y = (1 + 3x - 18x²)/(9x + 3), from part (a). Method: divide the fraction out into a linear part plus a small remainder. That exposes BOTH asymptotes at once and also makes differentiating easy. Step 1: Divide 1 + 3x - 18x² by 9x + 3. Write the numerator in descending powers first: -18x² + 3x + 1. -18x² ÷ 9x = -2x; -2x(9x + 3) = -18x² - 6x; subtracting leaves 3x + 1 + 6x = 9x + 1 9x ÷ 9x = 1; 1(9x + 3) = 9x + 3; subtracting leaves 1 - 3 = -2 So y = -2x + 1 - 2/(9x + 3) Step 2: Vertical asymptote where the denominator is zero. 9x + 3 = 0 x = -1/3 Step 3: Oblique asymptote. As x → ±∞ the term 2/(9x + 3) → 0, so y → -2x + 1. Step 4: Stationary points. Differentiate the divided form, writing the last term as -2(9x + 3)⁻¹ and using the chain rule. dy/dx = -2 - 2 × (-1)(9x + 3)⁻² × 9 = -2 + 18/(9x + 3)² Set dy/dx = 0: 18/(9x + 3)² = 2 (9x + 3)² = 9 9x + 3 = 3 or 9x + 3 = -3 9x = 0 or 9x = -6 x = 0 or x = -2/3 Step 5: Find the y-values from y = -2x + 1 - 2/(9x + 3). At x = 0: 9(0) + 3 = 3, so y = 0 + 1 - 2/3 = 1/3. Point (0, 1/3). At x = -2/3: 9(-2/3) + 3 = -6 + 3 = -3, so y = 4/3 + 1 - 2/(-3) = 4/3 + 1 + 2/3 = 3. Point (-2/3, 3). Step 6: Extra guide points for the sketch. y = 0 when the numerator is zero: 1 + 3x - 18x² = 0, i.e. 18x² - 3x - 1 = 0 Factorise (product 18 × (-1) = -18, sum -3, so split -3x as -6x + 3x): 18x² - 6x + 3x - 1 = 6x(3x - 1) + (3x - 1) = (6x + 1)(3x - 1) = 0 x = -1/6 or x = 1/3. Step 7: Which side of the oblique asymptote each branch lies, from the sign of -2/(9x + 3). For x < -1/3 the denominator is negative, so the term is positive and the left branch lies ABOVE y = -2x + 1; y → +∞ as x → (-1/3)⁻. For x > -1/3 the term is negative, so the right branch lies BELOW y = -2x + 1; y → -∞ as x → (-1/3)⁺. Sketch: two branches separated by x = -1/3. Left branch: comes in above y = -2x + 1 from the far left, dips to a MINIMUM at (-2/3, 3), then rises to +∞ at x = (-1/3)⁻. It never meets the x-axis, since its lowest point has y = 3. Right branch: rises from -∞ at x = (-1/3)⁺, crosses the x-axis at x = -1/6, reaches a MAXIMUM at (0, 1/3), crosses back at x = 1/3, and then falls away below y = -2x + 1. Asymptotes: x = -1/3 and y = -2x + 1. Stationary points: (-2/3, 3) and (0, 1/3).
More worked questions on this topic
- The curve y=(ax²+bx+c)/(x+d) has vertical asymptote x=−1 and oblique asymptote y=2x+2, with c≠2 (3 marks)
- A function has f(x) = x + 4 for -4 <= x < -1 and f(x) = x^2 + 2 for -1 <= x < 2, with f(x) = f( (3 marks)
- For y = f(x) with f(x) = 1 - sqrt(q^2 - x^2), q > 1 (a semicircle), sketch y = 1/f(x), giving a (3 marks)
- For the recurrence u_{n+1}=(−2u_n+6)/(u_n−1), sketch y=(−2x+6)/(x−1)−x, giving the asymptote eq (3 marks)
- Region R is enclosed by the curve y=3x²/[(x+1)(3x²+x+1)], the line x=4 and the x-axis. For x≥0, (3 marks)
More A-Level H2 Maths topics
Functions · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · Vector geometry in three dimensions · all of A-Level H2 Maths