Write down the coordinates of M.
About this question
Question 4.1 of Mathematics Grade 12 Paper 1, NSC Examination (May/June) (DBE, May/June 2025). 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 graph of \(f(x) = \frac{4}{x - 3} + 4\) is drawn. M is the point where the asymptotes of \(f\) intersect. C and D are the \(x\)- and \(y\)-intercepts of \(f\) respectively. A is the point on \(f\) that is closest to M.
Diagram: a hyperbola with a vertical dashed asymptote to the right of the \(y\)-axis and a horizontal dashed asymptote above the \(x\)-axis, meeting at M. The left branch comes in from the left just below the horizontal asymptote, cuts the positive \(y\)-axis at D, cuts the positive \(x\)-axis at C and then drops steeply down next to the vertical asymptote. The right branch lies above and to the right of M; A is marked on this branch at the point nearest to M.
Asymptotes:
\[x = 3 \text{ and } y = 4\]\(M(3\,;\,4)\)
More Questions On Functions and Inverse Functions
\(f(x) - g(x) \gt 0\) means \(f(x) \gt g(x)\): the parabola is above the line outside the intersection points P (\(x = 0\)) and E (\(x = 3\)).
\(x \lt 0\) or \(x \gt 3\)
The \(y\)-intercept is \(7\).
\(y = -x + 7\)
To the right of \(x = 3\), the branch of \(g\) is below or on \(f\) up to the intersection \(D(5\,;\,6)\). To the left of \(x = 3\), \(g\) lies above \(y = 8\), so above \(f\).
\(3 \lt x \leq 5\)
C is the \(y\)-intercept of \(f\):
\[f(0) = 0 - 0 - 5\]\(y = -5\)
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.
\(y = -\sqrt{-\frac{x}{6}}\), \(x \leq 0\)
Maximum where:
\[\frac{d(EH)}{dx} = -2x + 2 = 0 \quad \therefore x = 1\]\[EH = -(1)^2 + 2(1) + 3 = 4\]Maximum length of \(EH = 4\) units