Cambridge O Level Additional Mathematics · Syllabus 4037 · Logarithmic and Exponential Functions
Asymptote
What is Asymptote?
A straight line that a curve approaches ever more closely without ever meeting it. For the exponential family y equals k times e to the power n x plus a, the asymptote is the horizontal line y equals a, provided k is not zero, because the term k times e to the power n x approaches zero but never reaches it. For the logarithmic family y equals k times the natural logarithm of the bracket a x plus b, the asymptote is the vertical line found by setting a x plus b equal to zero, because the logarithm is undefined there and grows without bound on the side where the bracket is positive.
This definition is part of the Logarithmic and Exponential Functions chapter in Cambridge O Level Additional Mathematics.
Asymptote in context
The exponential function \(e^{x}\) and the natural logarithm \(\ln x\) are inverse functions, so \(\ln(e^{x})=x\) for every real \(x\) and \(e^{\ln x}=x\) for every positive \(x\). Their graphs are reflections of each other in the line \(y=x\), and their domains and ranges exchange: \(e^{x}\) accepts every real number and produces only positive values, with the horizontal asymptote \(y=0\), while \(\ln x\) accepts only positive numbers and produces every real value, with the vertical asymptote \(x=0\). That inverse relationship is why \(\ln\) is the tool that removes \(e\), and \(e\) is the tool that removes \(\ln\).
A logarithm is an exponent. Writing \(\log_a b\) asks one question and one question only: what power of \(a\) gives \(b\)? Everything in this chapter — the graphs, the asymptotes, the four laws, the equations — is a consequence of that single sentence and of the conditions that keep it meaningful.
Common mistakes with Asymptote
- Moving a vertical translation into the exponent: reading \(3e^{-2x}+4\) as \(3e^{-2x+4}\). Why it fails The \(+4\) is applied after the exponential has been evaluated, so it raises every point — and the asymptote — by \(4\). Inside the exponent it would instead multiply the curve by the constant \(e^{4}\): \(3e^{-2x+4}=3e^{4}e^{-2x}\), whose asymptote is still \(y=0\) and whose \(y\)-intercept is \(3e^{4}\approx163.8\), not \(7\). Do this Evaluate at \(x=0\) as a check: \(3e^{0}+4=7\) tells you immediately which reading you have.
- Confusing horizontal and vertical asymptotes: giving \(x=4\) as the asymptote of \(y=3e^{-2x}+4\), or \(y=2\) as the asymptote of \(y=2\ln(3x-6)\). Why it fails An exponential is bounded in the \(y\)-direction and unbounded in \(x\), so its asymptote must be horizontal. A logarithm is the reverse. The directions are exchanged precisely because the functions are inverses. Do this Ask which variable is restricted. Restricted \(y\) means a horizontal asymptote; restricted \(x\) means a vertical one.
- Saying an exponential graph “reaches” or “touches” its asymptote. Why it fails It would contradict the range. \(e^{x}>0\) for every real \(x\); there is no value of \(x\) at which \(e^{x}=0\), however large and negative \(x\) becomes. \(e^{-100}\approx3.7\times10^{-44}\), and still not zero. Do this Use “approaches”, “tends to” or “gets arbitrarily close to”, and never draw the curve meeting the line.
Questions students ask about Asymptote
What is the asymptote of \(y=ke^{nx}+a\), and of \(y=k\ln(ax+b)\)?
For \(y=ke^{nx}+a\) the asymptote is the horizontal line \(y=a\), because \(ke^{nx}\) approaches zero but never reaches it, so the curve is bounded in \(y\) and unbounded in \(x\). For \(y=k\ln(ax+b)\) the asymptote is the vertical line found from \(ax+b=0\), because the logarithm is undefined there; the domain is \(ax+b>0\). An exponential graph never touches its asymptote: \(e^{x}>0\) for every real \(x\).