Write down the range of \(f\).
About this question
Question 4.5 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.
The asymptote of \(f\) passes through \(D(1\,;\,4)\), so it is \(y = 4\), and \(f\) lies below it.
\(y \lt 4\), i.e. \(y \in (-\infty\,;\,4)\)
This question also appeared in
- NSC May/June 2025 P1, Question 6.2 (1 marks)
- NSC May/June 2025 P2, Question 7.1 (1 marks)
- NSC November 2025 P1, Question 5.2 (1 marks)
- NSC May/June 2024 P1, Question 6.1 (1 marks)
- NSC November 2024 P1, Question 4.2 (1 marks)
- NSC May/June 2021 P1, Question 5.3 (1 marks)
- NSC November 2021 P1, Question 7.3 (1 marks)
- NSC November 2020 P1, Question 4.2.2 (1 marks)
More Questions On Functions and Inverse Functions
Let \(f(x) = 0\):
\[\frac{-6}{x + 2} - 1 = 0\]\[-6 = x + 2\]\[x = -8\]\(A(-8\,;\,0)\)
\(A(-6\,;\,-6)\), \(B(0\,;\,6)\)
\[AB = \sqrt{(0 - (-6))^2 + (6 - (-6))^2}\]\[AB = \sqrt{36 + 144} = \sqrt{180}\]\(AB = 6\sqrt{5} \approx 13{,}42\) units
The minimum value of \(f(x)\) is \(-9\) (at D).
\(f(x) - k \gt 0\) for all \(x\) requires the whole graph shifted down by \(k\) to stay above the \(x\)-axis.
\[-9 - k \gt 0\]\(k \lt -9\)
For the maximum:
\[D'(x) = -2x + 1 = 0 \text{, so } x = \frac{1}{2}\](or \(x = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}\))
\[D\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 6\]Maximum distance \(= \frac{25}{4} = 6{,}25\) units
The minimum turning point is \(T(1\,;\,-4)\).
\(y \geq -4\), or \(y \in [-4\,;\,\infty)\)
- Horizontal asymptote: \(y = -1\) (dotted line).
- \(y\)-intercept: \(f(0) = 1 - 1 = 0\), so the graph passes through the origin \((0\,;\,0)\), which is also the only \(x\)-intercept.
- Shape: since \(0 \lt a \lt 1\) the graph is decreasing; it is large for negative \(x\) and gets closer to \(y = -1\) as \(x\) increases.
A decreasing exponential curve through \((0\,;\,0)\), approaching the asymptote \(y = -1\) on the right.