Question 8
Determine the equation of the asymptote of the function $f(x)=\log_9(x-3)-4$.
Not quite. The $-4$ moves the graph down 4 units. Moving a graph up or down slides it along a vertical asymptote, so it can’t create one at $x=-4$. In 2022, 11.9% of Queensland students chose (A).
Not quite. The asymptote is where the inside of the log reaches 0. $x-3=0$ gives $x=3$, not $-3$. Replacing $x$ with $x-3$ shifts the graph 3 units right. In 2022, 10.4% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2022, 68.1% of Queensland students chose (C).
Not quite. At $x=4$, $x-3=1$ and $\log_9 1=0$, so $f(4)=-4$. The graph passes through $(4,-4)$, so it crosses the line $x=4$ instead of approaching it. In 2022, 8.9% of Queensland students chose (D).
Near the bottom, the gap between the curve and the dashed line is narrower than a pixel. Five units below the blue point, the curve is only $9^{-5}\approx0.00002$ units from the asymptote.
This is $y=\log_9 x$. You can only take the log of a positive number, so the curve plunges down beside the $y$-axis, $x=0$, without ever touching it.
Moving the curve up or down slides it along the dashed line. The asymptote stays at .
Moving the curve sideways carries the asymptote with it. It is now at , where the inside of the log is 0.
This is the question’s function. The curve plunges down beside $x=3$ but never reaches it, so that is the asymptote. The blue point $(4,-4)$ is on the curve, so $x=4$ can’t be an asymptote.
You can only take the log of a positive number. Here the log is of $x-3$, so
$\displaystyle x-3\gt0$$\displaystyle \Rightarrow\ x\gt3$
The graph only exists to the right of $x=3$.
As $x$ gets close to 3 from the right, $x-3$ gets close to 0 and $\log_9(x-3)$ heads towards $-\infty$. The graph plunges down beside the line where the inside is zero.
$\displaystyle x-3=0$$\displaystyle \Rightarrow\ x=3$
The $-4$ moves every point down 4 units. A vertical line slid up or down stays where it is, so the asymptote is still $x=3$.
Check the other options: at $x=4$, $f(4)=\log_9 1-4=-4$, so the graph passes through $(4,-4)$. The values $-3$ and $-4$ are to the left of 3, where the function isn’t defined at all.
$\displaystyle x-3\gt0$$\displaystyle \Rightarrow\ x\gt3$
$\displaystyle x-3=0$$\displaystyle \Rightarrow\ x=3$
The log is undefined where its input is 0, and the vertical shift of $-4$ doesn’t move a vertical asymptote. The asymptote is $x=3$.
How Queensland students answered in 2022
From the QCAA Mathematical Methods subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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