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Question Video: Graphing Linear Functions by Plotting Points Mathematics

Consider the linear function 𝑓(π‘₯) = 5 βˆ’ 2π‘₯. We can draw a straight line to represent this function. Complete the table to find the coordinates of points on the line. Which of the following is the graph of the function? [A] Graph A [B] Graph B [C] Graph C [D] Graph D

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Video Transcript

Consider the linear function 𝑓 of π‘₯ is equal to five minus two π‘₯. We can draw a straight line to represent this function. Complete the table to find the coordinates of points on the line. Which of the following is the graph of the function?

There is also a third part to this question that we will consider later. In this question, we are given the linear function 𝑓 of π‘₯, which is equal to five minus two π‘₯. We recall that the values of π‘₯ are the inputs of the function and the values of 𝑓 of π‘₯ are the outputs. The first part of this question asks us to complete the table to find the coordinates of five points that lie on the line. These have π‘₯-values or inputs of negative two, negative one, zero, one, and two. And we need to find the corresponding values of 𝑓 of π‘₯.

Beginning with the first column, we have 𝑓 of negative two is equal to five minus two multiplied by negative two. This is equal to five plus four, which in turn is equal to nine. When π‘₯ is equal to negative two, 𝑓 of π‘₯ is equal to nine. Substituting π‘₯ is equal to negative one into our function gives us 𝑓 of negative one is equal to five minus two multiplied by negative one. This is equal to five plus two, which equals seven. The second missing value in our table is seven. When π‘₯ is equal to negative one, 𝑓 of π‘₯ is equal to seven. Repeating this when π‘₯ is equal to zero, we see that 𝑓 of zero is equal to five.

The third point that lies on the line has coordinates zero, five. 𝑓 of one is equal to three as five minus two multiplied by one is three. We have a fourth point with coordinates one, three. And finally, when π‘₯ is equal to two, 𝑓 of π‘₯ is equal to one. The five missing values from the table are nine, seven, five, three, and one. And this means that the points with coordinates negative two, nine; negative one, seven; zero, five; one, three; and two, one all lie on our straight line.

The second part of this question asks us to match one of the graphs with our function. The easiest way to do this is to consider the five points we have just found. All four lines pass through the point with coordinates zero, five. This means that they have a 𝑦-intercept equal to five. However, the only line that passes through the points negative two, nine; negative one, seven; one, three; and two, one is graph (C). We can therefore conclude that this is the graph of the function 𝑓 of π‘₯ which is equal to five minus two π‘₯.

Whilst it was not required to answer this question, it is worth noting that our straight line has 𝑦-intercept equal to five and slope or gradient equal to negative two. This corresponds to the straight line with equation 𝑦 is equal to negative two π‘₯ plus five, where our equation is written in slope–intercept form, 𝑦 equals π‘šπ‘₯ plus 𝑏. As this equation matches the function in this question, we know that our answer is correct.

The last part of this question says the following. Which of these points is not on the line? Is it (A) negative four, 13, (B) negative three, 11, (C) three, negative one, (D) four, negative three, or (E) six, negative six?

There are a few ways we could answer this question. Firstly, we could simply extend our table to contain the π‘₯-values negative three, negative four, positive three, positive four, and six. Alternatively, we could extend the straight line on our graph and see which points it passes through. However, there is an easier way to answer this question. We notice that all the values of 𝑓 of π‘₯, our 𝑦-coordinates, are odd. We can actually go one stage further. The line with equation 𝑦 is equal to negative two π‘₯ plus five, or 𝑦 is equal to five minus two π‘₯, will only have 𝑦-coordinates which are odd. This means that it could not pass through the point with coordinates six, negative six. The correct answer to this part of the question is option (E).

We can check this by working out the value of 𝑓 of six. Substituting π‘₯ equals six into our function gives us five minus two multiplied by six. This is equal to negative seven and not negative six. The straight line will therefore pass through the point with coordinates six, negative seven, but not the point with coordinates six, negative six.

We have now answered all three parts to the question. We have completed the table to find the coordinates of points that lie on the line. We have found the graph that corresponds to the function, and we have identified a point that does not lie on the line.

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