Video: Pack 4 β€’ Paper 1 β€’ Question 17

Pack 4 β€’ Paper 1 β€’ Question 17

04:55

Video Transcript

Consider the graphs below. Each graph shows a proportionality relationship. Fill in the table below by matching each graph to its corresponding statement.

The first proportionality relationship is 𝑦 is directly proportional to π‘₯, which we can write like this. And this is equivalent to the equation 𝑦 is equal to π‘˜π‘₯, where π‘˜ is a constant. Now, we notice that this equation is in the form 𝑦 is equal to π‘šπ‘₯ plus 𝑐, where π‘š is equal to our constant π‘˜ and 𝑐 is equal to zero.

So we’re looking for a straight line graph, which has a gradient of π‘˜. And since 𝑐 which is the 𝑦-intercept is equal to zero, this graph must pass through the origin. And so we spot that this must be graph D since graph D is a straight line graph passing through the origin. And so we can put this in our table.

The next proportionality relationship is 𝑦 is proportional to the square of π‘₯, which we can write as 𝑦 is directly proportional to π‘₯ squared. And this gives us the equation 𝑦 is equal to π‘˜π‘₯ squared, where π‘˜ is a constant.

Now, let’s see what happens to the sign of the 𝑦-values when we substitute in a positive and negative value of π‘₯. At π‘₯ equals one, 𝑦 is equal to π‘˜. And at π‘₯ is equal to negative one, 𝑦 is also equal to π‘˜. And so what this tells us is that for both positive and negative values of π‘₯, the sign of the 𝑦-values will remain the same.

And so looking at our graphs, we can see that this must be graph A or graph C since for both positive and negative π‘₯-values, the sign of the 𝑦-values remains the same.

Next, let’s substitute π‘₯ equals zero into our equation. This gives us that 𝑦 is also equal to zero. And so our graph must pass through the origin. And since graph C does pass through the origin, but graph A does not, that means that this proportionality relationship must match to graph C. And so we can put this in our table.

The next proportionality relationship says that 𝑦 is inversely proportional to π‘₯. And this means that 𝑦 is directly proportional to one over π‘₯ or the inverse of π‘₯. And this is equivalent to the equation 𝑦 is equal to π‘˜ over π‘₯.

Now, we notice that if we substitute π‘₯ equals zero into this equation, we’re dividing π‘˜ by zero. And since anything divided by zero is undefined, this means that at π‘₯ equals zero, 𝑦 is undefined. This tells us that our graph does not have a value at π‘₯ equals zero. And we see that this must relate to either graph A or graph B since both graphs C and D pass through the origin.

Now, let’s see what happens when we substitute in positive and negative π‘₯-values. At π‘₯ equals one, 𝑦 equals π‘˜. And at π‘₯ is equal to negative one, 𝑦 is equal to negative π‘˜. And so this tells us that when the sign of the π‘₯-values changes, the sign of the 𝑦-values also changes.

And looking at graph B for the positive π‘₯-values, we have positive 𝑦-values. And for the negative π‘₯-values, we have negative 𝑦-values. And so on this graph, when the sign of the π‘₯-value changes, the sign of the 𝑦-values also changes. And so this proportionality relationship must match to graph B.

The final proportionality relationship is 𝑦 is inversely proportional to the square of π‘₯. And this tells us the 𝑦 is directly proportional to one over π‘₯ squared. And this proportionality relationship gives us the equation 𝑦 is equal to π‘˜ over π‘₯ squared, where π‘˜ is again a constant.

Now, although the only graph remaining is graph A, let’s just check that that matches to this proportionality relationship just to be sure. Similarly to the previous proportionality relationship, if we substitute π‘₯ equals zero into this equation, we see that 𝑦 is undefined. And so therefore, this must be graph A or graph B.

Now, we’ll substitute in positive and negative π‘₯-values to see what happens to the sign of the 𝑦-values. At π‘₯ equals one, 𝑦 is equal to π‘˜ and at π‘₯ is equal to negative one, 𝑦 is also equal to π‘˜. And so therefore, when the sign of the π‘₯-value changes, the sign of the 𝑦-values remains the same. And we can see that this happens in graph A since for both the positive and negative π‘₯-values, the 𝑦-values have the same sign.

So therefore, we can complete our table by matching the final proportionality relationship to graph A. And so this is our final answer to the question.

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