Write down the coordinates of A.
About this question
Question 4.2 of Mathematics Grade 12 Paper 1, NSC Examination (DBE, November 2025). It tests the Term 1 ATP topic Exponential and Logarithmic Functions.
- TermTerm 1
- ATP topicExponential and Logarithmic Functions
- PaperPaper 1
- Marks1 mark for the method (formula and substitution) and the correct final answer
The graph of \(f(x) = \log_{\frac{1}{3}} x\) is drawn. Point A is the \(x\)-intercept of \(f\) and \((3\,;\,t)\) lies on \(f\).
Diagram: a decreasing logarithmic curve that lies to the right of the \(y\)-axis, with the \(y\)-axis as its vertical asymptote. It comes down steeply close to the \(y\)-axis, cuts the \(x\)-axis at A, and continues to decrease slowly below the \(x\)-axis, passing through the point \((3\,;\,t)\).
At the \(x\)-intercept:
\[\log_{\frac{1}{3}} x = 0\]\[x = \left(\frac{1}{3}\right)^0 = 1\]\(A(1\,;\,0)\)
This question also appeared in
- NSC November 2020 P1, Question 4.2.1 (2 marks)
- NSC November 2020 P1, Question 5.1 (1 marks)
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So \((4\,;\,3)\) lies on \(h\).
\(h\) is decreasing and always above its asymptote \(y = 0\).
\(0 \lt y \lt 3\)
Swap \(x\) and \(y\):
\[x = \log_{\frac{1}{3}} y\]\(y = \left(\frac{1}{3}\right)^x\) (or \(y = 3^{-x}\))
Since \(\left(\frac{1}{3}\right)^{-1} = 3\):
\(t = -1\)
The asymptote is \(y = -16\), so \(q = -16\).
\[f(x) = 2^{x + p} - 16\]Substitute \(T(3\,;\,16)\):
\[16 = 2^{3 + p} - 16\]\[2^{3 + p} = 32 = 2^5\]\[3 + p = 5\]\(p = 2\) and \(q = -16\)