N2020 P2 Q1

N2020 P2 Q1

Junior College 2
5 marks
Free

A quadratic curve has its minimum point at \((1,-2)\) and has gradient \(5\) at the point where \(x=2\). Find the equation of the curve.[5]

Default solution

With minimum \((1,-2)\), write \(y=a(x-1)^2-2\), where \(a>0\).

\(\frac{\mathrm dy}{\mathrm dx}=2a(x-1)\). At \(x=2\), \(2a=5\), so \(a=\frac52\).

Therefore \(y=\frac52(x-1)^2-2\).

Answer:\(y=\frac52(x-1)^2-2\)

Need help? Join our JC Math tuition classes.

Learn more