Video: AQA GCSE Mathematics Higher Tier Pack 2 • Paper 3 • Question 27

Sara has made a catapult for a school project. She first loads the catapult and then fires it. Before the catapult is fired, the projectile’s height can be modelled using the equation 𝑦 = 0 for 0 ≤ 𝑥 < 1. After the catapult is fired, the projectile’s height can be modelled using the equation 𝑦 = (𝑥 − 1)(9 − 𝑥) for 1 i≤ 𝑥 ≤ 9. 𝑦 is the height of the projectile in meters and 𝑥 is the time in seconds. (a) Draw a graph to show the height of the projectile between 0 and 9 seconds. (b) Work out the average rate of change in height between the times 𝑥 = 2 and 𝑥 = 6. You may find it helpful to draw a connecting line between the two corresponding points on the graph.

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Video Transcript

Sara has made a catapult for a school project. She first loads the catapult and then fires it. Before the catapult is fired, the projectile’s height can be modelled using the equation: 𝑦 equals zero for zero is less than or equal to 𝑥 which is less than one. After the catapult is fired, the projectile’s height can be modelled using the equation: 𝑦 equals 𝑥 minus one multiplied by nine minus 𝑥 for one is less than or equal to 𝑥 which is less than or equal to nine. 𝑦 is the height of the projectile in meters and 𝑥 is the time in seconds. Part a) Draw a graph to show the height of the projectile between zero and nine seconds.

There is also a part b to this question which we’ll look at in a moment. So the height of this projectile can be modelled using these two equations which describe how the height of projectile is changing during different parts of their first nine seconds. We’re told that 𝑦 is the height of the projectile in meters and 𝑥 is the time in seconds. So we can go ahead and add those two labels onto the set of coordinates axes we’ve been given.

We’re also told that Sara first loads the catapult. So this first equation where the height is always equal to zero must be the period during which she’s loading that catapult because the height is not changing. She then fires the catapult when 𝑥 is equal to one. That is, after a time of one second because this is the point at which the height of the projectile starts changing.

Now, we’re going to need a little bit of space in order to do some working out to draw this graph. So I’m going to clear some of the information off the screen. So here are those two equations. And remember, we’re drawing this graph for the whole nine seconds. Let’s begin by looking at the scales that we’ve been given on the two axes. On the horizontal axis, we see that 10 small squares represents the time of five seconds. By dividing by five, we see that two small squares then represents one second on the horizontal axis. On the vertical axis, we see that five small squares represents five meters. Again, dividing by five, we see that one square on the vertical axis represents one meter.

Let’s begin then with this first equation which, remember, describes the height of the projectile while the catapult is being loaded. The 𝑦-value or the height of the projectile is equal to zero throughout this first second. Now, remember that two small squares on the horizontal axis represents one second. So this first part of the journey can be represented using a horizontal line along the 𝑥-axis, that’s where 𝑦 is equal to zero, between the 𝑥-values of zero and one.

Now let’s consider this second equation which describes the height of the projectile once the catapult has actually been fired. We have two linear brackets in 𝑥 multiplied together which means the result will be a quadratic expression in 𝑥. So we’re going to be sketching a quadratic curve. We note, first of all, that the coefficient of 𝑥 squared will be negative. It is negative one because we have 𝑥 multiplied by negative 𝑥 to give negative 𝑥 squared. This means that we’ll be sketching a negative parabola, sometimes called an n-shaped or a sad parabola because we have a negative coefficient of 𝑥 squared.

Next, we consider the points at which this parabola will intersect with the 𝑥-axis, which are the points where 𝑦 is equal to zero because 𝑦 is equal to zero everywhere on the 𝑥-axis. Setting 𝑦 equal to zero gives the equation 𝑥 minus one multiplied by nine minus 𝑥 is equal to zero. We solve this equation in exactly the same way we usually solve a factorized quadratic. Which is, we set each bracket in turn equal to zero giving 𝑥 minus one equals zero and nine minus 𝑥 equals zero.

This first equation can be solved by adding one to each side, giving 𝑥 is equal to one. This tells us that one of the roots of this quadratic equation is one. And so our graph intersects the 𝑥-axis at a value of one. This makes sense because we’ve already drawn this horizontal part of the graph here which extends to an 𝑥-value of one. And if our graph is to be continuous, it needs to continue on from this point. The second equation can be solved by adding 𝑥 to each side, giving nine equals 𝑥 or 𝑥 equals nine. This means that the second root of our quadratic is nine. And remembering that two small squares on the horizontal axis represent one second, we can draw this point two seconds to the left of the value of 10 on our 𝑥-axis.

So we found the roots of this quadratic and we know its general shape. But we need another feature in order to draw it accurately. Let’s find the coordinates of its turning point, that is the point at the very top of this graph. As quadratic graphs are symmetrical either side of their turning point, the 𝑥-coordinate of this point will be the average of the two 𝑥-coordinates of the roots. That’s the average of one and nine, one plus nine over two. One plus nine is 10 and dividing by two gives five. To find the 𝑦-coordinate of this turning point, we substitute 𝑥 equals five into the equation of the curve, giving 𝑦 equals five minus one multiplied by nine minus five. That’s four multiplied by four which is 16.

So the graph has a turning point at the point five, 16 which we can plot. Remember, one small square represents one meter on the vertical axis. So the turning point of our graph is at this point here.

Now, we could find some more points that lie on this curve by substituting other values of 𝑥, for example, 𝑥 equals two, three, and so on. But, with the two roots and the coordinates of the turning point, we’ve got enough information to sketch this curve reasonably well. We join the points with a smooth symmetrical curve. So there is our answer to part a. We’ve drawn a graph to show the height of this projectile during the first nine seconds. The graph is horizontal between 𝑥-values of zero and one. And we then have a negative parabola between 𝑥-values of one and nine with a turning point at the point five, 16.

Part b) of the question says: Work out the average rate of change in height between the times 𝑥 equals two and 𝑥 equals six. You may find it helpful to draw a connecting line between the two corresponding points on the graph.

This graph is a graph of height against time. So it is a form of distance-time graph. When we’re asked to work out a rate of change of a distance-time graph, this means we’re looking at its gradient. In this case, we’re asked for an average rate of change between two times because the graph is a curve. And so its gradient and its instantaneous rate of change are always changing. The hint is that we may find it helpful to draw a line that connects the two points with 𝑥-values of two and six. So let’s use this hint. Remember, two small squares on a horizontal axis represent one second. So the values of two and six are two small squares right of one and five, respectively.

To work out the average rate of change in height between these times then, we’re looking for the gradient of the straight line that joins these two points on the graph. The gradient of a straight line is defined as the change in 𝑦 divided by the change in 𝑥. And if we think of two points on the line as having coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, we can formalize this as a 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. We know our 𝑥-values are six and two. So we just need to work out the corresponding 𝑦-values. And there are two ways that we can do this.

One way is to read the values directly from our graph. For example, when 𝑥 is equal to two, we can go horizontally across from our curve to the 𝑦-axis. And we see that the 𝑦-value here is seven. The other method will be to substitute each 𝑥-value into the equation of the curve at this point. So when 𝑥 is equal two, we have that 𝑦 is equal to two minus one multiplied by nine minus two. That’s one multiplied by seven which is again equal to seven.

Using either of these two methods, we see that the 𝑦-value when 𝑥 is equal to six is 15. Either of these two methods will be fine. Now, if you read the values from your graph, you may have got slightly different values, for example, 7.5 or eight for the 𝑦-value when 𝑥 is equal to two. This is okay as long as your graph was reasonably accurate to begin with. The 𝑦-values that you read off will be sufficiently accurate for you to use in your calculation of the gradient. So substituting these values into our calculation of the gradient gives 15 minus seven over six minus two. 15 minus seven is eight and six minus two is four. And eight divided by four is two.

We were dividing a quantity measured in meters by a quantity measured in seconds, which means that the units for this gradient, the units for the average rate of change, are meters per seconds. We found then that the gradient and therefore the average rate of change in height between the times 𝑥 equals two and 𝑥 equals six is two meters per second. And in the first part of this question, we drew a graph to represent the height of this projectile between zero and nine seconds.

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