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SingaporeAdditional MathematicsSyllabus dot point

What do the graphs of exponential and logarithmic functions look like, and how are they related?

Sketch graphs of exponential and logarithmic functions, identify their key features, and recognise them as reflections of each other

A focused answer to the N(A)-Level Additional Mathematics outcome on exponential and logarithmic graphs. Their shapes, intercepts and asymptotes, and how each is the reflection of the other in the line y equals x.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to recognise and sketch the graphs of exponential functions y=axy = a^x and logarithmic functions y=logaxy = \log_a x, to mark their key features (intercepts and asymptotes), and to understand that the two are inverses, so each graph is the reflection of the other in the line y=xy = x. Knowing the shapes lets you interpret growth and decay and check answers from the equation-solving topics.

The answer

The exponential graph y equals a to the x

For a base a>1a > 1, the graph of y=axy = a^x:

  • passes through (0, 1)(0,\ 1), because a0=1a^0 = 1,
  • is always positive and increasing, rising more and more steeply as xx grows,
  • has the xx-axis as a horizontal asymptote (y=0y = 0): the curve gets ever closer to it for large negative xx but never touches it.

If 0<a<10 < a < 1 the curve instead decreases (models decay), still passing through (0,1)(0, 1) with the same asymptote y=0y = 0.

The logarithmic graph y equals log of x

For base a>1a > 1, the graph of y=logaxy = \log_a x:

  • passes through (1, 0)(1,\ 0), because loga1=0\log_a 1 = 0,
  • is increasing but flattens out, growing more and more slowly,
  • has the yy-axis as a vertical asymptote (x=0x = 0): the curve plunges downward as xx approaches 00 from the right,
  • is defined only for x>0x > 0, because you cannot take the log of zero or a negative number.

The reflection relationship

Because y=logaxy = \log_a x is the inverse of y=axy = a^x, the two graphs are mirror images in the line y=xy = x. This swaps their features neatly: the exponential's yy-intercept (0,1)(0, 1) becomes the log's xx-intercept (1,0)(1, 0), and the exponential's horizontal asymptote y=0y = 0 becomes the log's vertical asymptote x=0x = 0.

Transformations

Adding a constant shifts the curve: y=ax+ky = a^x + k moves the exponential up by kk, which also moves its asymptote to y=ky = k. Recognising a shift helps you read graphs in modelling questions.

Examples in context

Example 1. Modelling growth and decay. An increasing exponential models a growing investment or population; a decreasing exponential (0<a<10 < a < 1) models radioactive decay or cooling. The shape of the curve, read off the graph, tells you at a glance whether a quantity is growing or shrinking and how fast.

Example 2. Checking an equation solution. When you solve 2x=202^x = 20 and get x4.32x \approx 4.32, a quick sketch of y=2xy = 2^x confirms the answer is sensible: the curve passes 1616 at x=4x = 4 and 3232 at x=5x = 5, so a value just above 44 is right. The graph is a fast sanity check on the algebra.

Try this

Q1. State the coordinates where y=5xy = 5^x crosses the yy-axis. [1 mark]

  • Cue. At (0,1)(0, 1), since 50=15^0 = 1.

Q2. Write down the equation of the asymptote of y=log2xy = \log_2 x and the domain of the function. [2 marks]

  • Cue. Vertical asymptote x=0x = 0; domain x>0x > 0.

Q3. Explain why the graphs of y=4xy = 4^x and y=log4xy = \log_4 x are reflections of each other in y=xy = x. [2 marks]

  • Cue. log4x\log_4 x is the inverse of 4x4^x, and the graph of an inverse is the reflection of the original in the line y=xy = x.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSketch the graph of y=2xy = 2^x, stating the coordinates of the point where it crosses the yy-axis and the equation of its asymptote.
Show worked answer →

The curve rises steeply for positive xx and approaches zero for negative xx.

It crosses the yy-axis at (0,1)(0, 1), because 20=12^0 = 1.

The horizontal asymptote is y=0y = 0 (the negative xx-axis), since 2x2^x never reaches zero.

What markers reward: an increasing curve through (0,1)(0, 1), the asymptote y=0y = 0 clearly marked, and the curve staying above the xx-axis everywhere.

Original3 marksState two features of the graph of y=log10xy = \log_{10} x, and explain how it is related to the graph of y=10xy = 10^x.
Show worked answer →

Features: it crosses the xx-axis at (1,0)(1, 0) (since log101=0\log_{10} 1 = 0), and it has a vertical asymptote at x=0x = 0; it is defined only for x>0x > 0.

Relation: y=log10xy = \log_{10} x is the inverse of y=10xy = 10^x, so its graph is the reflection of y=10xy = 10^x in the line y=xy = x.

What markers reward: the intercept (1,0)(1, 0), the vertical asymptote x=0x = 0 with the domain x>0x > 0, and identifying the log graph as the reflection of the exponential graph in y=xy = x.

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