# Question 4:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $PQ = -3$ | B1 | If simultaneous equations used, scored for $m = -3$ |
| Attempts equation of $l$, e.g. $y - 13 = -3(x+2)$ | M1 | Using their gradient and one point; sight of embedded values sufficient; perpendicular gradient gives M0 |
| $y = -3x + 7$ | A1 | Condone $m = -3,\ c = 7$ |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to use minimum $(4, -5)$, e.g. $y = \ldots(x-4)^2 - 5$ | M1 | Way One: $y = A(x-4)^2 - 5$ (condone $A=1$); Way Two: $y = ax^2+bx+c$ with valid method |
| Attempts to use $(-2, 13)$ with $y = a(x-4)^2 - 5 \Rightarrow a = \ldots$ | dM1 | Do not award for only substituting minimum into linear equation |
| $y = \frac{1}{2}(x-4)^2 - 5$ or $y = \frac{1}{2}x^2 - 4x + 3$ | A1 | |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Two of: $y > -3x+7$, $\ y < \frac{1}{2}(x-4)^2 - 5$, $\ x < -2$ | M1 | |
| All three: $y > -3x+7$, $\ y < \frac{1}{2}(x-4)^2 - 5$, $\ x < -2$ | A1 | |

# Question dM1 (Part b - Quadratic):

**Way One** | dM1 | Attempts to use $(-2,13)$ with $y=a(x-4)^2-5 \Rightarrow a=...$

**Way Two** | dM1 | Attempts to use $(-2,13)$ and $(4,-5)$ in $y=ax^2+bx+c$ using two equations: $4a-2b+c=13$ and $16a+4b+c=-5$, with $2a\times4+b=0$ to find $a$, $b$ and $c$

$y=\frac{1}{2}(x-4)^2-5$ or equivalent e.g. $y=\frac{1}{2}x^2-4x+3$ or $2y=x^2-8x+6$ | A1 | Cannot be $C=...$

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# Question (c):

**M1 criteria:** Two of $y>-3x+7$, $y<\frac{1}{2}(x-4)^2-5$, $x<-2$ | M1 | Ignore any others; line from (a) must have negative gradient, curve from (b) must be positive quadratic. Allow $x<k$ where $k$ is any value between $-2$ and $4$. Note: $-3x+7 < y < \frac{1}{2}(x-4)^2-5$ also scores M1. But $-3x+7 < x < \frac{1}{2}(x-4)^2-5$ with one other inequality is M0.

All three of $y>-3x+7$, $y<\frac{1}{2}(x-4)^2-5$, $x<-2$ (and no others) | A1 | Allow all three with $x<k$ where $k$ between $-2$ and $4$. If set notation used, must use $\cap$ not $\cup$. Allow consistent use of $>\leftrightarrow\geq$ for all inequalities including $x<-2$.

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