Why is \(f\) a function?
About this question
Question 4.7 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
- Marks1 mark
Sketched below are the graphs of \(f(x) = -a^x + q\) and \(g(x) = -x^2 + 2x + 3\). B and C are the \(x\)-intercepts of \(g\). D, the turning point of \(g\), is also a point on the asymptote of \(f\). T is a point on \(f\) such that \(DT\) is parallel to the \(y\)-axis.
Diagram: \(g\) is a parabola opening downwards, cutting the \(x\)-axis at B (left of O) and C (right of O), with its turning point D in the first quadrant. A horizontal dashed line through D is the asymptote of \(f\). \(f\) is a decreasing exponential curve that lies below this asymptote: it approaches the asymptote far to the left, passes through the \(y\)-axis at the same point as \(g\), crosses \(g\) again between D and C, and cuts the \(x\)-axis to the right of C. A vertical dashed line from D down to \(f\) meets \(f\) at T.
For every \(x\)-value there is exactly one corresponding \(y\)-value (\(f\) is a one-to-one relation; any vertical line cuts it only once).
More Questions On Functions and Inverse Functions
\(B(0\,;\,6)\), \(C(6\,;\,0)\) (C is the \(x\)-intercept of \(g^{-1}\): \(\frac{1}{2}x - 3 = 0\))
\[BC = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}\]\(AB = AC = \sqrt{180}\) (\(g\) and \(g^{-1}\) are symmetric about \(y = x\)), so \(\triangle ABC\) is isosceles.
Midpoint of BC: \(M(3\,;\,3)\), so \(AM \perp BC\) is the height.
\[AM = \sqrt{(3 + 6)^2 + (3 + 6)^2} = \sqrt{162} = 9\sqrt{2}\]\[\text{Area} = \frac{1}{2} \times BC \times AM = \frac{1}{2}(6\sqrt{2})(9\sqrt{2})\]Area of \(\triangle ABC = 54\) square units
\(x = 3 \text{ or } x = -1\), so \(B(-1\,;\,0)\) and \(C(3\,;\,0)\)
The vertical asymptote of \(f\) is \(x = -p\) and it meets \(g\) on the \(x\)-axis.
\((-p\,;\,0)\)
\(f(x) + g(x) \geq 0\) means \(f(x) \geq -g(x)\), where \(-g(x) = -\frac{1}{2}x - 4\) (the reflection of \(g\) in the \(x\)-axis).
\(-g\) also passes through \(A(-8\,;\,0)\) and through \((0\,;\,-4) = B\), so \(f\) and \(-g\) meet at \(x = -8\) and \(x = 0\).
\(f\) lies on or above \(-g\) from A up to the vertical asymptote, and from B onwards.
\(-8 \leq x \lt -2 \text{ or } x \geq 0\)
So \(C\left(-3\,;\,\frac{19}{8}\right)\).
Reflecting in \(y = x\) swaps the coordinates.
\(C'\left(\frac{19}{8}\,;\,-3\right)\)