Question 4
Consider the function $f(x)=\log_p(x+q)$ where $p\gt1$ and $0\lt q\lt1$.
Which of the following could be the graph of $f(x)$?
Not quite. (A) passes through the origin, so $f(0)=\log_p(q)=0$, which means $q=1$. But $q$ must be less than 1.
Not quite. The asymptote of (B) is to the right of the $y$-axis, so the graph never meets the $y$-axis. The asymptote is $x=-q$, so that would need $q\lt0$.
That’s it. Step through the hints below to see why.
Not quite. (D) crosses the $y$-axis above the origin, so $\log_p(q)\gt0$, which needs $q\gt1$. Its asymptote is also to the left of $x=-1$.
The shaded strip is $-1\lt x\lt0$, where the asymptote has to be when $0\lt q\lt1$.
With $q\lt0$ the asymptote $x=-q$ is to the right of the $y$-axis, so the graph never meets the $y$-axis. That breaks $0\lt q\lt1$.
With $q=0$ the asymptote is the $y$-axis itself, so the graph never meets it. That breaks $0\lt q\lt1$.
This is allowed. The asymptote $x=-q$ is inside the strip, and $f(0)=\log_p(q)$ is negative, so the graph crosses the $y$-axis below the origin. The $x$-intercept, $1-q$, is between 0 and 1.
$q=1$ isn’t allowed. $f(0)=\log_p(1)=0$, so the graph passes through the origin, and the asymptote $x=-1$ is on the edge of the strip.
With $q\gt1$, $f(0)=\log_p(q)$ is positive, so the graph crosses the $y$-axis above the origin, and the asymptote is to the left of the strip. That breaks $0\lt q\lt1$.
With these values the curve is very close to graph ().
You can only take the log of a positive number, so $x+q\gt0$. The vertical asymptote is where $x+q=0$.
$$x=-q$$
With $0\lt q\lt1$, the asymptote is between $x=-1$ and $x=0$: just to the left of the $y$-axis. That rules out (B).
The $y$-intercept is $f(0)$.
$$f(0)=\log_p(q)$$
With $p\gt1$, the log of a number less than 1 is negative. So the graph crosses the $y$-axis below the origin. That rules out (A), which goes through the origin, and (D), which crosses above it.
The $x$-intercept is where $x+q=1$, so $x=1-q$, between 0 and 1.
Only (C) has all three features: an asymptote just left of the $y$-axis, a negative $y$-intercept, and an $x$-intercept between 0 and 1.
You don’t need $p$. Changing $p$ stretches the graph vertically, but the asymptote and the $x$-intercept don’t move, and the sign of $f(0)$ stays the same. Try the $p$ slider in the figure.
Asymptote: $x+q=0$, so $x=-q$, and $-1\lt-q\lt0$.
$y$-intercept: $f(0)=\log_p(q)\lt0$, because $p\gt1$ and $0\lt q\lt1$.
$x$-intercept: $x+q=1$, so $x=1-q$, between 0 and 1.
Only graph (C) has an asymptote just left of the $y$-axis and crosses the $y$-axis below the origin.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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