Video: Pack 3 β€’ Paper 2 β€’ Question 18

Pack 3 β€’ Paper 2 β€’ Question 18

05:39

Video Transcript

Sketch the graph of 𝑦 is equal to sin of π‘₯ degrees for zero is less than or equal to π‘₯, which is less than or equal to 360.

Let’s think about what we know about the trigonometric graph 𝑦 is equal to sin of π‘₯ degrees. First, we know that it’s a periodic graph. That means it repeats and then it does so every 360 degrees. We also know that it has a maximum 𝑦-value of one and a minimum 𝑦-value of negative one. Finally, it’s useful to remember that the sin curve passes through the 𝑦-axis at zero, whereas the cos graph passes through at one.

As long as we remember these facts and the general shape which is a wave, the next part is just to sketch it. The sin curve has rotational symmetry about the point 180 degrees zero. The graph of 𝑦 is equal to sin π‘₯ is as shown.

Remember since the question says to sketch the graph, it doesn’t need to be perfectly to scale. It should, however, include all the relevant points of intersection with the axes and the turning point should be roughly in the right place. It’s also useful to know what to do if we’re asked to sketch the graph over a different domain.

In this case, the domain is π‘₯ has got to be greater than or equal to zero, but less than or equal to 360 degrees. If this domain was larger, for example, up to 720 degrees, we’d remember that the graph is periodic. It repeats every 360 degrees. So we can continue the wave shape that we already have.

The graph of 𝑦 is equal to 𝑓 of π‘₯ is shown on both grids below.

Part one, On the first grid, draw the graph of 𝑦 is equal to a half 𝑓 of π‘₯.

Let’s recall the notation. When we see a function 𝑦 is equal to 𝑓 of π‘₯, that just means that it’s an equation for 𝑦 in terms of π‘₯. 𝑓 of π‘₯ could really be any function such as π‘₯ squared or four π‘₯ cubed. Since we don’t know the exact equation though, we just write 𝑦 is equal to 𝑓 of π‘₯.

For a function 𝑓 of π‘₯, we can perform several transformations. π‘Ž multiplied by 𝑓 of π‘₯ is a stretch in the 𝑦-direction by a scale factor of π‘Ž, whereas 𝑓 of 𝑏 multiplied by π‘₯ is a stretch in the π‘₯-direction by a scale factor of one over 𝑏.

Let’s compare these transformations, which are both structures, to the equation that we’ve been given. 𝑦 is equal to a half of 𝑓 of π‘₯ looks just like the first transformation function. In this case, multiplying a whole function by π‘Ž lead to a stretch in the 𝑦-direction by a scale factor of π‘Ž. That means then that 𝑦 is equal to a half multiplied by 𝑓 of π‘₯ is a stretch in the 𝑦-direction by a scale factor of one-half.

Let’s look at what happens to each of our coordinates. Halving the value of the output, which is the 𝑦-coordinate, means the coordinate negative four, two becomes negative four, one. The coordinate negative two, two becomes negative two, one. The coordinate negative one, zero stays the same since half of zero is still zero. Zero, negative two becomes zero, negative one. One, negative two becomes one, negative one. Two, negative one becomes two, negative a half. And three, negative one becomes three, negative a half.

We can now see that our function 𝑓 of π‘₯ has been stretched in the 𝑦-direction by a scale factor of one-half. It looks like it’s been compressed.

Part two, On the second grid, draw the graph of 𝑦 is equal to 𝑓 of π‘₯ plus two.

For the function 𝑓 of π‘₯, the graph of 𝑓 of π‘₯ plus 𝑐 is a translation by the vector zero, 𝑐. And the graph of 𝑓 of π‘₯ plus 𝑑 is a translation by the vector negative 𝑑, zero. Once again, we need to compare the function we’ve been given with the transformations listed.

In this case, something is being added to the π‘₯ before the function is applied. In our list, that looks like 𝑓 of π‘₯ plus 𝑑, which leads to a translation by the vector negative 𝑑, zero. That means our graph is going to be translated by the vector negative two, zero. That’s two units left. When we move this first coordinate two units left, it goes from negative four, two to negative six, two. The coordinate negative two, two goes to negative four, two and so on.

The graph of 𝑦 is equal to 𝑓 of π‘₯ plus two has been translated two units to the left. It’s also worth noting that there are two further transformations that we need to be able to apply: negative 𝑓 of π‘₯ is a reflection in the π‘₯-axis and 𝑓 of negative π‘₯ is a reflection in the 𝑦-axis.

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