Calculate the coordinates of the point of intersection of \(g\) and \(h\).
About this question
Question 5.4 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
- Marks3 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.
\((6\,;\,7)\)
More Questions On Functions and Inverse Functions
\(C\left(0\,;\,\frac{9}{2}\right)\), \(D(-5\,;\,2)\), \(E(-5\,;\,0)\)
OCDE is a trapezium with parallel sides \(DE = 2\) and \(OC = 4{,}5\) and height \(OE = 5\).
\[\text{Area} = \frac{1}{2}(DE + OC) \times OE\]\[= \frac{1}{2}(2 + 4{,}5)(5)\]\(\text{Area} = \frac{65}{4} = 16{,}25\) square units
\(g \parallel AE\), so \(m_g = \frac{1}{4}\)
\[y - (-9) = \frac{1}{4}\left(x - (-3)\right)\]\[y = \frac{1}{4}x + \frac{3}{4} - 9 = \frac{1}{4}x - \frac{33}{4}\]Equate \(g\) and \(f\):
\[x^2 - 2x - 24 = \frac{1}{4}x - \frac{33}{4}\]\[4x^2 - 8x - 96 = x - 33\]\[4x^2 - 9x - 63 = 0\]\[(4x - 21)(x + 3) = 0\]\[x = \frac{21}{4} \text{ or } x = -3\](\(x = -3\) is K)
\(x_R = \frac{21}{4} = 5{,}25\)
C lies on the asymptote of \(g\), so \(q = -5\)
Substitute \(D(2\,;\,-9)\) into \(g(x) = a \cdot 2^x - 5\):
\[-9 = a(2)^2 - 5\]\[-4 = 4a\]\(a = -1\,;\; q = -5\)
\(D(1\,;\,4)\)
Swap \(x\) and \(y\):
\[x = \frac{2}{3}y + 2\]\[\frac{2}{3}y = x - 2\]\(g^{-1}: y = \frac{3}{2}x - 3\)
So \(h\) is the reflection of \(f\) in the \(y\)-axis. \(A(5\,;\,6)\) maps to \(A'(-5\,;\,6)\).
\[AA' = 5 - (-5)\]\(AA' = 10\) units