Calculate the coordinates of A.
About this question
Question 5.2.2 of Mathematics Grade 12 Paper 1, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 1 ATP topic Functions and Inverse Functions.
- TermTerm 1
- ATP topicFunctions and Inverse Functions
- PaperPaper 1
- Marks2 marks for the method (formula and substitution) and the correct final answer
The graphs of \(f(x) = \frac{-6}{x + 2} - 1\) and \(g(x) = \frac{1}{2}x + 4\) are drawn below. A and B are the \(x\)- and \(y\)-intercepts of \(f\) respectively. A is also a point of intersection of \(f\) and \(g\).
Diagram: \(f\) is a hyperbola with a vertical dashed asymptote to the left of the \(y\)-axis and a horizontal dashed asymptote just below the \(x\)-axis. The left branch of \(f\) rises steeply as it nears the vertical asymptote and cuts the \(x\)-axis at A. The right branch lies below the \(x\)-axis, cutting the \(y\)-axis at B and flattening out towards the horizontal asymptote. \(g\) is a straight line with a positive gradient through A, crossing the left branch of \(f\) again and cutting the positive \(y\)-axis.
Let \(f(x) = 0\):
\[\frac{-6}{x + 2} - 1 = 0\]\[-6 = x + 2\]\[x = -8\]\(A(-8\,;\,0)\)
Other points that earn marks
- Also accepted: A lies on \(g\), so \(\frac{1}{2}x + 4 = 0\) gives \(x = -8\).
This question also appeared in
- NSC May/June 2026 P2, Question 3.1 (3 marks)
- NSC May/June 2025 P1, Question 4.5 (3 marks)
- NSC November 2021 P1, Question 10.2 (3 marks)
More Questions On Functions and Inverse Functions
Substitute \(B\left(2\,;\,-\frac{5}{9}\right)\) into \(f(x) = a^x - 1\):
\[-\frac{5}{9} = a^2 - 1\]\[a^2 = \frac{4}{9}\]\(a = \frac{2}{3}\) (since \(a \gt 0\))
Swap \(x\) and \(y\):
\[x = \frac{1}{2}y + \frac{9}{2}\]\(y = 2x - 9\)
\(k\) is parallel to \(g\), so the gradient of the tangent is 1.
\[f'(x) = 2x - 2 = 1 \text{, so } x = \frac{3}{2}\]\[f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 2\left(\frac{3}{2}\right) - 3 = -\frac{15}{4}\]So the point of contact is \(\left(\frac{3}{2}\,;\,-\frac{15}{4}\right)\).
\[k(x) = x - 3 - n\]\[-\frac{15}{4} = \frac{3}{2} - 3 - n\]\(n = \frac{9}{4} = 2{,}25\)
Swap \(x\) and \(y\):
\[x = -6y^2\]\[y^2 = -\frac{x}{6}\]\[y = \pm\sqrt{-\frac{x}{6}}\]Since \(f^{-1}(x) \leq 0\), take the negative root.
\(y = -\sqrt{-\frac{x}{6}}\), \(x \leq 0\)
Turning point of \(f\): \((1\,;\,18)\)
Turning point of \(h\): \((2\,;\,15)\)
The graph moves 1 unit right and 3 units down.
\(f(x + p)\) shifts the graph \(p\) units left, so:
\[1 - p = 2 \text{, so } p = -1\]\[18 + q = 15 \text{, so } q = -3\]\(p = -1\) and \(q = -3\)