# Daily Maths Revision – Week 17 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

This week, we’re exploring:

### Question 1: Plans and Elevations

Plans and elevations are 2-dimensional representations of 3-dimensional shapes. The plan view is how the shape looks from directly above. The front and side elevations are how the shape looks from directly in front and from the side. You can think of the plans and elevations as shadows of the shape.

Draw the plan view, front elevation and side elevation of the shape below. The arrow shows the front view.

From above, you won’t be able to tell how tall the shape is – you will only be able to see 2 squares. From the front, you will see a pile of three squares with an extra one at the right on the bottom. From the side, you will see a pile of three squares. Using a ruler, we can draw these views accurately.

### Question 2: Drawing Quadratic Graphs from a Table

When asked to draw a quadratic graph, you will usually be given a table to complete. Unlike the linear graphs we looked at in week 11, these graphs won’t be straight lines. Instead, they will be a smooth curve called a parabola.

Let’s use the table below to draw the graph with equation 𝑦 = 𝑥2 + 2𝑥 – 3.

We simply need to substitute each value of 𝑥 into the equation 𝑦 = 𝑥2 + 2𝑥 – 3. Be particularly careful with negative signs. If you are putting the values into a calculator, put brackets around any negative values.

𝑥 = -3:            𝑦 = (-3)2 + 2 × (-3) – 3
= 9 – 6 – 3
= 0

𝑥 = 0:              𝑦 = (0)2 + 2 × (0) – 3
= 0 + 0 – 3
= -3

𝑥 = 3:              𝑦 = (3)2 + 2 × (3) – 3
= 9 + 6 – 3
= 12

Now, we can plot these points on a set of axes.

Finally, we can join the coordinates. Make sure you draw a smooth curve rather than lots of short, straight lines. This includes at the bottom of the graph – it continues to curve as it changes direction.

### Question 3: Rotation

When rotating a shape, you will be given a centre of rotation, an angle and a direction. The only exception to this is if you are rotating through 180°; in this case, the direction is irrelevant.

You should always ask for tracing paper when rotating – it makes the task much easier. If working at home, baking paper makes a good alternative.

Let’s rotate this shape 90° anticlockwise about the point (2, 4).

First, we need to identify the centre of rotation – the point (2, 4). Then, we place our tracing paper over the image and draw both the shape and the point.

Now, use a pencil to hold the point in the same place and rotate the shape through 90° anticlockwise.

Finally, remove the tracing paper and draw the shape in the new position. When you’ve drawn the shape, you should check the answer by putting the tracing paper back.

If asked to describe a transformation, you can identify it as a rotation if the shape has turned. In some cases, however, this could also be a reflection so make sure you consider both. You then need to specify that it’s a rotation and give the angle and direction.

Describe the single transformation that maps shape A onto shape B.

The easiest to find the centre of rotation is by trial and improvement. We can draw shape A and try a few different rotations until we find the one that maps to B.

Let’s start by trying the point (4, 3). Draw shape A on your tracing paper and rotate it until your shape has the same orientation as shape B.

As you can see, the shape is too far to the left and too far down. Let’s try a point a little further up – say (4, 4).

Now the orange shape overlays shape B so this is the correct centre of rotation. Now, we just need to write down what we did.

A rotation, centre (4, 4), 90° clockwise.

### Question 4: The Equations of Parallel Lines

As we saw in week 14, any straight line can be written in the form 𝑦 = 𝑚𝑥 + 𝑐 where 𝑚 is the gradient of the line and 𝑐 is the 𝑦-intercept. This week, we aren’t given the gradient of the line. Instead, we are given the equation of a line that is parallel.

If two lines are parallel, then their gradients are the same. So, if we can identify the gradient in one line, we have the gradient of the other line. In some questions, you need to find the gradient using two coordinates first. You will have covered this in week 8 so have a look back if you need a reminder.

A straight line is parallel to the line with equation 𝑦 = 7𝑥 – 5 and passes through the point (2, 3). Work out the equation of the line in the form 𝑦 = 𝑚𝑥 + 𝑐.

The gradient of the line with equation 𝑦 = 7𝑥 – 5 is the coefficient of 𝑥, 7.  Putting this in the form 𝑦 = 𝑚𝑥 + 𝑐 gives:

𝑦 = 7𝑥 + 𝑐

Since the line passes through the point (2, 3), we can substitute these coordinates in to find the value of 𝑐. Remember, coordinates are in the format (𝑥, 𝑦) so 𝑥 = 2 and 𝑦 = 3.

𝑦 = 7𝑥 + 𝑐
3 = 7 × 2 + 𝑐
3 = 14 + 𝑐
𝑐 = -11

Putting this into our equation gives us our final answer: 𝑦 = 7𝑥 – 11.

### Question 5: The Difference of Two Squares

The difference of two squares is a specific type of quadratic. Most quadratics can be arranged into the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐. The difference of two squares is recognisable because the middle term is missing – more specifically they are in the form 𝑚2 – 𝑛2. They are one square subtracted from another.

Factorise 𝑥2 – 1.

When we had expressions in the form 𝑥2 + 𝑏𝑥 + 𝑐, we looked for a pair of numbers that multiplied together to give c and added together to give b.

We can do the same here. In this case, 𝑐 = -1 and 𝑏 = 0. The only numbers that multiply to give -1 and add together to give 0 are 1 and -1. So:

𝑥2 – 1 = (𝑥 + 1)(𝑥 – 1)

We can see that the two brackets are almost identical, just the sign in the middle differs. This is the key to the difference of two squares: 𝑚2 – 𝑛2 will always factorise to (𝑚 + 𝑛)(𝑚 – 𝑛).

For example:   𝑥2 – 16 = 𝑥2 – 42
= (𝑥 + 4)(𝑥 – 4)

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