(a) (i) Complete the table of values for \( y = x^2 + x – 4 \).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | 8 | 2 | -2 | -4 | -4 | -2 | 2 | 8 |
(ii) On the grid, draw the graph of \( y = x^2 + x – 4 \) for \( -4 \leq x \leq 3 \).
(b) (i) Write down the co-ordinates of the lowest point of the graph.
(ii) Write down the equation of the line of symmetry of the graph.
(c) Use your graph to solve the equation \( x^2 + x – 4 = -3 \).
(d)
(i) In the diagram, a line L has been drawn on a 1 cm2 grid.
Write down the equation of the line L.
Give your answer in the form \( y = mx + c \).
(ii) Find the area of the shaded triangle.
▶️ Answer/Explanation
(a) (i) Ans: See table above.
Substitute each \( x \) value into \( y = x^2 + x – 4 \). For example, when \( x = -4 \), \( y = (-4)^2 + (-4) – 4 = 16 – 4 – 4 = 8 \). Repeat for all \( x \) values.
(ii) Ans: Correctly plotted points and smooth correct curve.
Plot the points from the table and draw a smooth parabola through them.
(b) (i) Ans: \((-0.5, -4.25)\).
The vertex (lowest point) of \( y = x^2 + x – 4 \) is at \( x = -\frac{b}{2a} = -0.5 \). Substituting gives \( y = -4.25 \).
(ii) Ans: \( x = -0.5 \).
The line of symmetry passes through the vertex, so it’s \( x = -0.5 \).
(c) Ans: \(-1.6 \leq x \leq -1.2\) and \(0.6 \leq x \leq 1.0\).
Find where the graph \( y = x^2 + x – 4 \) intersects \( y = -3 \). The solutions are the \( x \)-values at these points.
(d) (i) Ans: \( y = 2x – 3 \).
The line passes through points \((0, -3)\) and \((1.5, 0)\), giving slope \( m = 2 \) and \( y \)-intercept \( c = -3 \).
(ii) Ans: 9.
The triangle has base 3 cm and height 6 cm, so area \( = \frac{1}{2} \times 3 \times 6 = 9 \).
The scale drawing shows the positions of three towns, \( R, S \) and \( T \), on a map.
\( RS \) and \( ST \) are straight roads between the towns.
The scale is \( 1 \) centimetre represents \( 8 \) kilometres.
(a) Work out the actual distance between \( R \) and \( S \).
(b) Another town, \( V \), is on a bearing of \( 163° \) from \( R \) and on a bearing of \( 215° \) from \( T \).
Mark the position of \( V \) on the map.
(c) A man cycles at a constant speed of \( 24 \) km/h along the straight road from \( S \) to \( T \).
After 1 hour and 50 minutes he stops at a café, \( C \).
Mark the position of \( C \) on the map.
(d) A hotel, \( H \), is on a bearing of \( 321° \) from \( R \).
Work out the bearing of \( R \) from \( H \).
(e) Write the scale \( 1 \) cm to \( 8 \) km in the form \( 1:n \).
▶️ Answer/Explanation
(a) Ans: 48 km
Measure \( RS \) on the map (e.g., 6 cm).
Actual distance = \( 6 \times 8 = 48 \) km.
(b)
Use a protractor to draw bearings \( 163° \) from \( R \) and \( 215° \) from \( T \).
The intersection point is \( V \).
(c) Ans: Position \( C \) marked 5.5 cm from \( S \) towards \( T \)
Convert 1 hour 50 minutes to 1.833 hours.
Distance = \( 24 \times 1.833 = 44 \) km.
Map distance = \( 44 \div 8 = 5.5 \) cm.
(d) Ans: 141°
Bearing of \( R \) from \( H \) is \( 321° – 180° = 141° \).
(e) Ans: 1:800,000
Convert 8 km to cm: \( 8 \times 1000 \times 100 = 800,000 \) cm.