## Question 9:

### Part (a) – Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = \pm ax^2 \pm bx \pm c$; $x=0, H=2 \Rightarrow c=2$ **and either** $x=20, H=0.8 \Rightarrow 0.8=400a+20b+2$ **or** $x=9, \frac{dH}{dx}=0 \Rightarrow 18a+b=0$ | M1 | 3.3 – Uses $H=\pm ax^2 \pm bx \pm c$, uses $x=0, H=2$ correctly to find $c$, and uses one further condition |
| All three conditions used: $c=2$, $0.8=400a+20b+2$, and $18a+b=0$ | dM1 | 3.1b – Depends on M1; must use both endpoint and derivative condition |
| Solves to find $a=..., b=...$ | ddM1 | 1.1b – Solves two equations simultaneously |
| $H = -0.03x^2 + 0.54x + 2$ | A1 | 2.2a – Correct equation |

**Notes:** Condone use of $y$ for $H$ for method marks. Model of form $H=x^2+ax+b$ or $H=-x^2+ax+b$ scores no marks. Points $(-2, 0.8)$ and $(18, 2)$ also lie on parabola by symmetry.

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### Part (a) – Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=9$ at max $\Rightarrow H = A \pm B(x-9)^2$ **and either** $x=0, H=2 \Rightarrow 2=A+81B$ **or** $x=20, H=0.8 \Rightarrow 0.8=A+121B$ | M1 | 3.3 |
| Both: $2=A+81B$ **and** $0.8=A+121B$ | dM1 | 3.1b |
| $2=A+81B,\ 0.8=A+121B \Rightarrow A=4.43,\ B=-0.03$ | ddM1 | 1.1b |
| $H = 4.43 - 0.03(x-9)^2$ | A1 | 2.2a |

**Possible alternative 3:**
$$H = A\bigl((x-9)^2 - 81\bigr) + B$$
$$x=0, H=2 \Rightarrow B=2;\ x=20, H=0.8 \Rightarrow 0.8=40A+B \Rightarrow A=-0.03$$
$$H = 2 - 0.03\bigl((x-9)^2-81\bigr)$$

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### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any suitable limitation e.g.: $H$ is unlikely to be a quadratic in $x$; wind may affect path; air resistance not considered; ball is not a particle; ground unlikely horizontal; ball may spin; path unlikely parabolic | B1 | 3.5b |

**Do not accept:** statements outside range of throw (e.g. "$H$ will become negative", "model not valid for all $x$"); single-word vague answers; "air resistance", "friction", "spin" alone without context linking to model.

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### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=16 \Rightarrow H = -0.03(16)^2 + 0.54(16) + 2 = ...$ | M1 | 3.4 – Substitutes $x=16$ into equation to obtain value of $H$ |
| $H=2.96$; so Chandra would **not** be able to catch the ball | A1 | 3.2a – Correct value and conclusion |

**Alternative:** $2.5 = 4.43 - 0.03(x-9)^2 \Rightarrow x = 9 + \frac{\sqrt{579}}{3} = 17.02...$ so Chandra cannot catch ball (M1 for substituting $H=2.5$ into quadratic to find $x$; A1 for $x \approx 17$ and correct conclusion).

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