# Question Video: Graphing Linear Functions by Making Tables Mathematics

By making a table of values, determine which of the following is the function represented by the graph shown. [A] 𝑓(𝑥) = 3𝑥 + 1 [B] 𝑓(𝑥) = 3𝑥 − 1 [C] 𝑓(𝑥) = −3𝑥 + 1 [D] 𝑓(𝑥) = −3𝑥 − 1 [E] 𝑓(𝑥) = (1/3 𝑥) − 1

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### Video Transcript

By making a table of values, determine which of the following is the function represented by the graph shown. Is it (A) 𝑓 of 𝑥 is equal to three 𝑥 plus one? (B) 𝑓 of 𝑥 is equal to three 𝑥 minus one. (C) 𝑓 of 𝑥 is equal to negative three 𝑥 plus one. (D) 𝑓 of 𝑥 is equal to negative three 𝑥 minus one. Or (E) 𝑓 of 𝑥 is equal to one-third 𝑥 minus one.

In this question, we are given five functions and are asked to determine which of them is represented by the graph shown. As the graph shown is a straight line, we know it is a linear function. And this can be written in the form 𝑓 of 𝑥 is equal to 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient of the graph and 𝑏 is the 𝑦-intercept. We are told in this question that we must begin by constructing a table of values. To do this, we will identify the coordinates of some points that lie on the graph. Whilst we could choose any points that lie on the line, it is sensible to choose those with integer 𝑥- and 𝑦-values. The points with coordinates one, two; zero, negative one; and negative one, negative four all lie on the line.

By listing the 𝑥-coordinates in ascending order, we can add the values to our table as shown. We can now substitute the 𝑥-values into each of our options to see if the coordinates satisfy the functions. We will begin with 𝑥 is equal to one. In option (A), 𝑓 of one is equal to three multiplied by one plus one. This is equal to four. Since this is not equal to two, the point one, two does not lie on the graph of the function 𝑓 of 𝑥 is equal to three 𝑥 plus one. And we can therefore rule out option (A). In option (B), 𝑓 of one is equal to three multiplied by one minus one. This is equal to two, so the point one, two does satisfy option (B). In option (C), 𝑓 of one is equal to negative three multiplied by one plus one, which is equal to negative two. As this is not equal to two, we can rule out option (C).

The same is true of option (D), where 𝑓 of one is equal to negative four. Likewise, in option (E), 𝑓 of one is equal to negative two-thirds, which is not equal to two. We have therefore ruled out options (A), (C), (D), and (E) as the point with coordinates one, two does not lie on the graphs of these functions.

Going back to option (B), we’ll now calculate 𝑓 of zero. This is equal to three multiplied by zero minus one. This is equal to negative one. So the point zero, negative one does lie on the graph of the function 𝑓 of 𝑥 is equal to three 𝑥 minus one. Repeating this for the third point, we see that 𝑓 of negative one is equal to three multiplied by negative one minus one, which is equal to negative four. We can therefore conclude that all three points lie on the graph of the function 𝑓 of 𝑥 is equal to three 𝑥 minus one, and the correct answer is option (B).

It is worth noting that we could’ve found this directly from the graph. Since the straight line intersects the 𝑦-axis at negative one, we know that the 𝑦-intercept is negative one. This rules out options (A) and (C), which have a 𝑦-intercept equal to positive one. The graph slopes upwards from left to right, which means it has a positive slope or gradient. This rules out option (D) and once again option (C). We could then calculate this slope by dividing the rise by the run between two points that lie on the line. The slope of our line is equal to three divided by one, which is equal to three, which rules out option (E). And this confirms that the correct function is option (B), 𝑓 of 𝑥 is equal to three 𝑥 minus one.