2026 TPSS P1 Q10

2026 TPSS P1 Q10

Secondary 4
5 marks

The expression \(x^2-8x+11\) is equivalent to \((x+a)^2+b\).

  1. Find the value of \(a\) and of \(b\).[2]
  2. Sketch the graph of \(y=x^2-8x+11\), indicating clearly the \(y\)-intercept and the turning point of the curve.[3]

Solution:

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Answer:(a) \(a=-4,\ b=-5\) (b) Upward-opening parabola with \(y\)-intercept \((0,11)\) and minimum \((4,-5)\).

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