In part (b) we are asked plot the points whose \(x\) and \(y\) values are recorded in the table above, and to draw a smooth curve through the points. The graph below uses a scale of 2 cm to represent 1 unit on the \(x\)-axis, and 1 cm to represent 1 unit on the \(y\)-axis.

A graph showing a smooth curve through the points in the table

For part (c), using the graph we estimate that the value of \(y\) when \(x = 3.5\) is approximately \(2.2\) (N.B. The exact answer is \(y = 2.25\)).

Using the graph to estimate the value of y at x = 3.5

Finally, in part (d) we make the following observations:

The equation of the axis of symmetry of the graph is \(x = 1\).

\(-4\) is the minimum value of the function \(y\).

When \(x^2 – 2x – 3 = 0\), i.e. \(y = 0\) then from the table we see that \(x = -1\) or \(x = 3\).

Updated on April 16th, 2012: Added some video explanations

In part (a), we given two triangles, \(JKL\) and \(MLP\), with \(JK\) parallel to \(ML\). \(LM = MP\), \(KLP\) is a straight line, angle \(JLM = 22^\circ\) and angle \(LMP = 36^\circ\).

Triangles JKL and MLP

Since \(LM\) and \(MP\) has the same length, it follows that \(\angle MLP = \angle MPL = \frac{180^\circ – 36^\circ}{2} = 72^\circ\).

\(\angle LJK = \angle JLM = 22^\circ\), since they are alternate angles.

\(\angle JKL = \angle MLP = 72^\circ\), since they are corresponding angles.

In part (b), we have a triangle, \(PQR\), and its image, \(P’Q'R’\).

A transformation of triangle PQR

The coordinates of \(P\) is \((2, 1)\) and the coordinates of \(Q\) is \((4, 3)\).

The transformation that maps triangle \(PQR\) onto triangle \(P’Q'R’\) is a reflection in the x-axis.

Under the translation \(\begin{pmatrix}3 \\ -6\end{pmatrix}\), P moves to the new point \(\begin{pmatrix}2 \\ 1\end{pmatrix} + \begin{pmatrix}3 \\ -6\end{pmatrix} = \begin{pmatrix}5 \\ -5\end{pmatrix}\), i.e. \((5, -5)\) and Q moves to the new point \(\begin{pmatrix}4 \\ 3\end{pmatrix} + \begin{pmatrix}3 \\ -6\end{pmatrix} = \begin{pmatrix}7 \\ -3\end{pmatrix}\), i.e. \((7, -3)\).

Updated on April 13th, 2012: Added some video explanations

In part (a) we are given an extract from a bus schedule (for details, check the image above). The bus begins its journey at Belleview, travels to Chagville and ends its journey at St. Andrews.

A bus and its travel route

Since the bus arrived in Chagville at 7:35 a.m. and departed at 7:45 a.m., the bus spent 10 minutes at Chagville.

The bus departed Belleview at 6:40 a.m. and arrived in Chagville at 7:35 a.m. Hence, it took the bus 55 minutes to travel from Belleview to Chagville.

We are now told that the bus travelled at an average speed of 54 km/hour from Belleview to Chagville. In other words, in 1 hour (or 60 minutes) the bus travelled 54 km. But the bus didn’t travel for 60 minutes, it travelled for 55 minutes. So, if we can compute how far the bus can travel in 1 minute going at an average speed of 54 km/hour then we’d be able to determine how far it would have travelled in the 55 minutes going at the same average speed. Now, in 1 minute the bus would have travelled \(\frac{54}{60} \text{ km} = \frac{9}{10} \text{ km}\). Therefore, in 55 minutes the bus would have travelled \(55 \times \frac{9}{10} \text{ km} = 49.5 \text{ km}\). Hence, the distance from Belleview to Chagville is 49.5 km.

In part (b), 4.8 litres of water is poured into a cylindrical bucket with a base area of \(300 \text{ cm}^2\). We have to find the height of the water in the bucket. Recall that the volume of a prism (a cylinder is a prism) is given by the product of its cross-sectional area times its perpendicular height.

A cylinder with water

In our case the cross-sectional area is given as the base area. Now, 4.8 litres is \(4800 \text{ cm}^3\) since 1 litre is \(1000 \text{ cm}^3\). Let \(h\) represent the height of the water in the bucket. Then,

For part (c), the area of the shaded face, in terms of \(h\), is \(4h \text{ cm}^2\). The volume of the cuboid, in terms of \(h\), is \(13 \text{ cm} \times 4h \text{ cm}^2 = 52h \text{ cm}^3\). If the volume of the cuboid is \(286 \text{ cm}^3\), then

You may notice that in the second equation we can solve for \(x\) in terms of \(y\) to get \(x = 2y – 1\). Substituting for \(x\) in the first equation gives us

And now, if \(x = 3\), then using the first or second equation we can solve for \(y\) to get \(y = 2\). Hence, \(x = 3\) and \(y = 2\) is the solution just as before.

Finally, in part (c) we are told that tickets for a football match are sold at $30 per adult and $15 per child and that a company bought 28 tickets.

If \(x\) of the tickets were for adults, then \(28 – x\) tickets were for children. It means then that \(\$30x\) was spent on tickets for adults and \(\$15(28 – x)\) was spent on tickets for children.

Therefore, the total amount spent on the 28 tickets is