Sketch exponential and logarithmic graphs
Connect eˣ and ln x as inverse functions, then identify domains, ranges, asymptotes and transformed graph features.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the range of y = eˣ?
What you will learn
- Sketch the basic graphs y = eˣ and y = ln x
- Use their inverse relationship and reflection in y = x
- Interpret domains, ranges and asymptotes for the syllabus graph forms
The basic graphs are inverses
The exponential function y = eˣ has domain all real x and range y > 0. It passes through (0, 1), increases, and approaches but never reaches the horizontal asymptote y = 0 as x decreases.
The natural logarithm y = ln x is its inverse: its domain is x > 0 and its range is all real y. It passes through (1, 0), increases, and has vertical asymptote x = 0.
Because inverse graphs swap x- and y-coordinates, y = ln x is the reflection of y = eˣ in y = x. The asymptote y = 0 of the exponential reflects to x = 0 for the logarithm.
Read transformations from the stated forms
For y = k eⁿˣ + a with k and n non-zero, the horizontal asymptote is y = a and the domain is all real x. If k > 0 the range is y > a; if k < 0 it is y < a. The sign of kn determines whether the graph increases or decreases.
For y = k ln(ax + b) with k and a non-zero, require ax + b > 0. Its vertical asymptote is ax + b = 0, or x = −b/a, and its range is all real numbers.
A labelled sketch should show the asymptote, an intercept or other reliable point, and the correct direction of the curve. A horizontal or vertical shift changes the asymptote; it does not remove it.
Pause and explain
Which graph is the inverse of y = eˣ?
Worked example
For y = 2eˣ − 3, state the domain, range, horizontal asymptote and y-intercept.
Show the worked solution
- The exponential is defined for every real x, so the domain is ℝ.
- Since eˣ > 0, 2eˣ − 3 > −3; the range is y > −3.
- The graph approaches y = −3 but never reaches it. At x = 0, y = 2(1) − 3 = −1.
Answer Domain ℝ; range y > −3; horizontal asymptote y = −3; y-intercept (0, −1).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For y = 3e²ˣ + 1, state its domain, range, asymptote and y-intercept.
Give me a hint
Use e²ˣ > 0 and substitute x = 0 for the intercept.
Compare my reasoning
- The exponent is defined for all real x, so the domain is ℝ.
- 3e²ˣ is positive, so y > 1; the horizontal asymptote is y = 1.
- At x = 0, y = 3e⁰ + 1 = 4.
Domain ℝ; range y > 1; asymptote y = 1; y-intercept (0, 4).
Look for these in your work
- Domain stated
- Range uses positive exponential
- Shifted asymptote and intercept found
For y = −2 ln(x + 4), give the domain, range and vertical asymptote.
Give me a hint
Set the logarithm argument greater than zero, then find where it equals zero.
Compare my reasoning
- x + 4 > 0, so x > −4.
- The logarithm can take every real value; multiplying by −2 preserves an all-real range.
- The argument is zero at x = −4, the vertical asymptote.
Domain x > −4; range ℝ; vertical asymptote x = −4.
Look for these in your work
- Log argument condition
- Range identified
- Asymptote located
A cooling sensor follows T(t) = 5e⁻ᵗ + 18 for t ≥ 0. Describe the starting value, long-term temperature and whether it rises or falls.
Give me a hint
Evaluate at t = 0, identify the horizontal asymptote, and use the negative exponent.
Compare my reasoning
- T(0) = 5 + 18 = 23.
- As t grows, e⁻ᵗ approaches zero, so T approaches 18.
- The exponential term decreases, so the temperature falls toward 18 without reaching the asymptote in this model.
It starts at 23, falls, and approaches the horizontal asymptote T = 18.
Look for these in your work
- Initial value evaluated
- Asymptote interpreted
- Direction justified
Common mistakes
- Giving y = 0 as the asymptote after a vertical shift
- Allowing a non-positive input to ln x
- Reflecting an inverse graph in an axis instead of y = x
- Stating the logarithm's range as positive only
What can you explain now?
What is the vertical asymptote of y = ln(2x − 6)?
Compare with the explanation
x = 3
The logarithm argument is zero at 2x − 6 = 0, and the domain is x > 3.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the domain and range of eˣ and ln x, and explain how their graphs are related.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.