Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 2 β€’ Question 29

Hannah draws this graph to identify the region 𝑅 represented by π‘₯ < βˆ’4, 𝑦 ≀ βˆ’2π‘₯ + 4, 𝑦 β‰₯ π‘₯ βˆ’ 2. Make two criticisms of her graph.

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Video Transcript

Hannah draws this graph to identify the region 𝑅 represented by π‘₯ is less than negative four, 𝑦 is less than or equal to negative two π‘₯ plus four, 𝑦 is greater than or equal to π‘₯ minus two. Make two criticisms of her graph.

Let’s begin by checking the equations against the lines drawn on the graph. Remember, a strict inequality, here that’s the first one π‘₯ is strictly less than negative four, is represented by a dotted line. We can see that the line π‘₯ is equal to negative four is in the correct place. A common mistake here is to think that it needs to be parallel to the π‘₯-axis whereas, in fact, π‘₯ is equal to negative four is perpendicular to the π‘₯-axis and passes through it at negative four. It is also dotted as we said was required.

The other two inequalities are not strict. And they should be represented by solid lines. We can see that the line 𝑦 is equal to negative two π‘₯ plus four is in the right place. It has a 𝑦-intercept of four and a gradient of negative two. However, it’s represented by a dotted line instead of a solid line. Our first criticism then is that the line 𝑦 is equal to negative two π‘₯ plus four is a dotted line when it should be solid because it’s less than or equal to negative two π‘₯ plus four. We will check the final line. It has a 𝑦-intercept of negative two and a gradient of one as we would expect from its equation. It’s also solid as required.

So the next thing we need to check is that the region 𝑅 is the correct region. To do this, we’re going to find one point that lies inside that region and check that the coordinates of that point satisfy each of the individual inequalities. A sensible starting point wherever possible is the point zero, zero. In this coordinate, π‘₯ is equal to zero and 𝑦 is equal to zero. And if we substitute zero into the first inequality, we get zero is less than negative four. This is a false statement. So in fact, we can see that she’s identified the wrong region.

In fact, we’re interested in everything on the other side of the line π‘₯ is equal to negative four. Let’s choose a point in the region I’ve labelled 𝑅 and see if that satisfies all three inequalities. Let’s choose the point negative six, zero. This time, the π‘₯-coordinate is negative six and the 𝑦-coordinate is zero. If we substitute π‘₯ is equal to negative six and 𝑦 is equal to zero into the second inequality, we get zero is less than or equal to negative two multiplied by negative six plus four.

A negative multiplied by a negative is positive. So negative two multiplied by negative six is positive 12. And zero is less than or equal to 12 plus four. Or zero is less than or equal to 16. This is of course true. We should also substitute π‘₯ is equal to negative six and 𝑦 is equal to zero into the third inequality. And doing so, we get zero is greater than or equal to negative six minus two or zero is greater than or equal to negative eight, which of course again is true. Our second criticism then is that she has shaded the wrong region. She should have shaded to the left of the line π‘₯ is equal to negative four as shown.

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