2024 GESS PRELIMS P1 Q1
2024 GESS PRELIMS P1 Q1
5 marks
The equation of a curve is \(y = -2x^2 + 3x + 5\).
- Write \(y = -2x^2 + 3x + 5\) in the form \(y = a(x - h)^2 + k\), where \(a\), \(h\) and \(k\) are constants.[3]
- Using your results in part (a), explain clearly why the maximum value of \(y\) is \(k\) when \(x = h\).[2]
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Answer:(a) \(y=-2\left(x-\frac34\right)^2+\frac{49}{8}\), so \(a=-2\), \(h=\frac34\), \(k=\frac{49}{8}\) (b) Maximum \(y=\frac{49}{8}=k\), attained when \(x=\frac34=h\)
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