Write down the length of \(OB\).
About this question
Question 8.2 of Mathematics Grade 12 Paper 1, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 2 ATP topic Differential Calculus.
- TermTerm 2
- ATP topicDifferential Calculus
- PaperPaper 1
- Marks1 mark
The graph of \(h(x) = -x^3 + px^2 + qx + r\) is drawn below. \(C(2\,;\,0)\) and \(D(5\,;\,0)\) are the \(x\)-intercepts and B is the \(y\)-intercept of \(h\). A and C are the turning points of \(h\).
Diagram: a cubic graph that comes down from the top left, cuts the positive \(y\)-axis at B, decreases to touch the \(x\)-axis at \(C(2\,;\,0)\) (a local minimum), rises to a local maximum A above the \(x\)-axis, then falls and cuts the \(x\)-axis at \(D(5\,;\,0)\) and continues downwards to the right.
B is the \(y\)-intercept: \(h(0) = r = 20\), so \(B(0\,;\,20)\).
\(OB = 20\) units
More Questions On Differential Calculus
\(f\) is increasing to the left of M and to the right of N.
\(x \lt -1\) or \(x \gt \frac{5}{3}\)
\(f\) decreases between the local maximum and the local minimum.
\(1 \lt x \lt \frac{5}{2}\)
\(f'(x) = -2x\)
Turning points where \(f'(x) = 0\): \(x = -\frac{1}{3}\) or \(x = 1\) (the \(x\)-intercepts of \(f'\))
\[f\left(-\frac{1}{3}\right) = \frac{1}{27} + \frac{1}{9} - \frac{1}{3} + 2 = \frac{49}{27}\]\[f(1) = -1 + 1 + 1 + 2 = 3\]\(\left(-\frac{1}{3}\,;\,\frac{49}{27}\right)\) and \((1\,;\,3)\)
Parallel to CD, so:
\[f'(x) = -1\]\[-2x - 6 = -1\]\[x = -\frac{5}{2}\]\[f\left(-\frac{5}{2}\right) = -\frac{25}{4} + 15 + 7 = \frac{63}{4}\]\(P\left(-\frac{5}{2}\,;\,\frac{63}{4}\right)\)