2025 Beatty Sec 3 EOY Q3

2025 Beatty Sec 3 EOY Q3

Secondary 3
5 marks
  1. The equation of a curve is \(y=a(x+b)^2+c\), where \(a\), \(b\) and \(c\) are constants. The solutions of \(a(x+b)^2+c=0\) on the graph are \(-2\) and \(8\), and the maximum value of \(y\) is \(3\). Find the values of \(a\), \(b\) and \(c\).[3]
  2. Using the values of \(a\), \(b\) and \(c\) found in (a), sketch the graph of \(y=a(x+b)^2+c\).[2]

Solution:

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Answer:(a) \(a=-0.12\), \(b=-3\), \(c=3\); (b) downward-opening parabola with vertex \((3,3)\), roots \(-2\), \(8\).

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