2020 H2 Maths Paper 2 Q1 — Worked Solution

Solution

5 marks

Since the curve is quadratic and has a minimum point at $(1, - 2)$ , we write it in vertex form,

y = a(x - 1)^{2} - 2,

where $a > 0$ is a constant to be found.

Differentiating,

\frac{dy}{dx} = 2a(x - 1).

Using the given gradient at $x = 2$ :

2a(2 - 1) = 5

2a = 5

a = \frac{5}{2}.

Substituting back,

y = \frac{5}{2}(x - 1)^{2} - 2

= \frac{5}{2}x^{2} - 5x + \frac{1}{2}.

To check your answer: at

x = 1

y = \frac{5}{2}(0)^{2} - 2 = - 2

, the stated minimum; and

\frac{dy}{dx} = 5(x - 1)

gives

5(1) = 5

x = 2

. ✓

$\therefore$ the equation of the curve is $y = \frac{5}{2}x^{2} - 5x + \frac{1}{2}$ (equivalently $y = \frac{5}{2}(x - 1)^{2} - 2$ ).