Exponential and Log Graphs (AA SL)
Exponential and logarithmic graphs share a distinctive curved shape and a single asymptote, and IB questions usually ask you to read off features - the asymptote, the intercept, the domain, or the range - rather than plot the whole curve. This page focuses on spotting those features quickly. It's part of the broader Exponentials & Logarithms topic.
11 questions on this sub-topic.
Reading the graph
Covered under IB syllabus reference SL2.9, which covers exponential and logarithmic functions and their graphs. For graph-reading questions, what actually matters is these two feature checks - neither is in the formula booklet, so you work them out from the equation each time.
Exponential: \(y = a\cdot b^{x} + k\)
Horizontal asymptote: \(y = k\)
As \(x\to-\infty\) (for \(b>1\)), \(b^x\to0\), so \(y\to k\) but never reaches it. The \(y\)-intercept is \(a+k\), found by substituting \(x=0\).
Logarithmic: \(y = \log_b(x)\)
Vertical asymptote: \(x = 0\); domain \(x>0\)
A log graph is the reflection of the matching exponential graph in the line \(y=x\), so its asymptote is vertical rather than horizontal, and it has no \(y\)-intercept.
Need the full syllabus wording and the log laws in detail? See Exponentials & Logarithms.
Worked examples
State the range of \(f(x)=3^{x}+4\).
Worked solution
An exponential is always positive: \(3^x>0\) for all \(x\). R1
Adding 4 gives \(f(x)>4\) (approached but never reached). M1
Range: \(f(x)>4\). A1
State the horizontal asymptote and \(y\)-intercept of \(y=2^{x}+3\).
(a)(i) State the horizontal asymptote.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) As \(x\to-\infty\), \(2^x\to0\), so \(y\to3\): asymptote \(y\). R1
\(=3\). A1
(a)(ii) \(y\)-intercept: \(2^0+3=4\Rightarrow(0,4)\). A1
The graph of \(y=e^{x}\) is reflected in the \(x\)-axis and then translated 5 units up to give the graph of \(g\).
(a) State the equation of \(g(x)\).
(b) State the horizontal asymptote of \(g\).
Worked solution
(a) Reflecting \(y=e^x\) in the \(x\)-axis gives \(y=-e^x\); translating this up 5 gives M1 \(g(x)=-e^x+5.\) A1
(b) As \(x\to-\infty,\ e^x\to0\), so \(g(x)\to5\): horizontal asymptote \(y=5.\) A1
Common mistakes
- Splitting \(\log(x+y)\) into \(\log x + \log y\). Logarithms don't distribute over addition or subtraction - a mistake that shows up when reading intercepts off a shifted log graph.
- Forgetting the domain of a logarithm. \(\log_a x\) is only defined for \(x>0\), so a log graph never crosses into negative \(x\)-values - stating a domain that includes \(x\le0\) is an automatic error.
- Stating the asymptote as an equation the curve reaches. \(y=k\) is a boundary the exponential graph gets arbitrarily close to but never touches - writing "\(y=k\) when \(x=-\infty\)" as though it's a coordinate is not accepted.
Ready to practise properly?
11 exponential-and-log-graph questions, marked instantly like the real exam.
Quick answers
What is the horizontal asymptote of an exponential graph?
For \(y=a\cdot b^x+k\), the horizontal asymptote is \(y=k\), since \(b^x\) approaches 0 as \(x\to-\infty\) (for \(b>1\)) but never actually reaches it.
Why does a log graph have a vertical asymptote instead of a horizontal one?
A log graph is the reflection of an exponential graph in the line \(y=x\), so the exponential's horizontal asymptote becomes a vertical one for the log graph, at the value of \(x\) excluded from the domain.