Question
Answer Key
Worked Solution
9758/2020/P2/Q01
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]
Equation of curve: $y=\frac{5}{2}(x-1)^2-2$
Consider this equation of the quadratic curve,
$$
y=a(x-h)^2+k
$$
At minimum point $(1,-2)$,
$$
\begin{array}{l}
y=a(x-1)^2-2 [B1]\\
\frac{\mathrm{d} y}{\mathrm{~d} x}=2 a(x-1)^{2-1}(1)+0 [B1] \\
\text { At } x=2, \frac{\mathrm{d} y}{\mathrm{~d} x}=5 \\
5=2 a(2-1) \\
a=\frac{5}{2}
\end{array}
$$
Equation of curve: $y=\frac{5}{2}(x-1)^2-2$
[A2]