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NSWMaths AdvancedQuick questions
Year 12: Functions
Quick questions on Exponential and logarithmic graphs: key features, transformations and inverse relationship
12short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.
What is the exponential graph?Show answer
For a general base $a > 0$, $a \neq 1$:
What is the inverse relationship?Show answer
$\ln$ is the inverse of $e^x$ on its domain:
What is transformations of $e^x$?Show answer
Apply the general transformation rules. For $y = a e^{b(x - h)} + k$:
What is transformations of $\ln x$?Show answer
For $y = a \ln(b(x - h)) + k$:
What is locating the asymptote of a shifted exponential?Show answer
Start with $y = e^x$, shift right by $1$: $y = e^{x - 1}$.
What is domain and asymptote of a logarithmic transformation?Show answer
Find the domain and vertical asymptote of $y = \ln(2 x - 6) + 1$.
What is using inverse to find a graph?Show answer
The inverse of $y = e^{x - 1} + 2$ is found by swapping $x$ and $y$ and solving: $x = e^{y - 1} + 2$ gives $y - 1 = \ln(x - 2)$, so $y = 1 + \ln(x - 2)$. Domain $(2, \infty)$, vertical asymptote $x = 2$. As expected, the asymptote of the original ($y = 2$) becomes the vertical asymptote of the inverse ($x = 2$).
What is wrong asymptote on a shifted exponential?Show answer
$y = a e^x + k$ has asymptote $y = k$, not $y = 0$. The horizontal asymptote moves with the vertical shift.
What is missing domain restriction on $\ln$?Show answer
$\ln(\text{negative}) = $ undefined. Always solve "argument $> 0$" before sketching.
What is confusing $\ln x$ with $\log_{10} x$?Show answer
In Maths Advanced, $\ln$ means natural log (base $e$) and $\log$ usually also means base $10$. Be careful when reading the question.
What is forgetting $e^x$ never reaches $0$?Show answer
$y = e^x > 0$ for every real $x$. The asymptote $y = 0$ is approached, not crossed.
What is sketching $\ln x$ as a power curve?Show answer
$\ln$ grows slowly, slower than any positive power of $x$. It also has unbounded negative values near $x = 0$, not a finite minimum.