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