# Question Video: Identifying the Graph of a Linear Function Using a Table of Values Mathematics • 6th Grade

By making a table of values, determine which of the following is the function represented by the graph shown. [A] 𝑦 = 5𝑥 + 1 [B] 𝑦 = (−1/5 𝑥) − 1 [C] 𝑦 = (−1/5 𝑥) + 1 [D] 𝑦 = (1/5 𝑥) − 1 [E] 𝑦 = (1/5 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) 𝑦 equals five 𝑥 plus one? (B) 𝑦 equals negative one-fifth 𝑥 minus one. Is it (C) 𝑦 equals negative one-fifth 𝑥 plus one? (D) 𝑦 equals one-fifth 𝑥 minus one. Or (E) 𝑦 equals one-fifth 𝑥 plus one.

Now, the question does actually tell us to make a table of values. So one thing that we can do is go through each of our functions and form a table of values and check whether those lie on the line given. Before we do though, we’re going to identify a handful of points that lie on the line given in our graph. Once we’ve done that, we’ll be able to quite quickly disregard the graphs that don’t match.

Since one small square represents one unit, our line passes through the point with coordinates five, two. Zero, one also lies on that line, as does negative five, zero. We’re now ready to create a table of values for each of the functions (A) through (E). Since the first number in each ordered pair represents the 𝑥-coordinate, we’re going to test the 𝑥-values of negative five, zero, and five. And we do so by substituting each of these values into each of our equations.

We begin with our first equation. When 𝑥 is equal to negative five, 𝑦 is five times negative five plus one, which is negative 24. Now, of course, our first coordinate is negative five, zero. So when 𝑥 is negative five, 𝑦 needs to be equal to zero. Since our value of 𝑦 is negative 24, we know that this point negative five, zero does not lie on the line 𝑦 equals five 𝑥 plus one. And so we’re not going to carry out the rest of the calculations. We’re going to move on to option (B).

This time, when 𝑥 is negative five, 𝑦 is negative one-fifth times negative five minus one. Well, that’s one minus one, which is equal to zero. So far so good then. The first coordinate is negative five, zero, which we’ve seen lies on our line. So let’s move on and check 𝑥 equals zero. Here, when 𝑥 equals zero, 𝑦 is negative one-fifth times zero minus one. And that’s negative one. But remember, our second coordinate had a 𝑦-value of one. And so this value of negative one tells us that our line cannot be 𝑦 equals negative one-fifth 𝑥 minus one. And we’re not going to worry about the third value in our table. We’re just going to move on to option (C).

This time, when 𝑥 is negative five, 𝑦 is negative one-fifth times negative five plus one. And that’s equal to two. Now, we know that we wanted our corresponding 𝑦-value to be zero. So we’re going to disregard option (C) and move on to (D). When 𝑥 is negative five, 𝑦 is a fifth times negative five minus one. And that’s negative two. Once again, this does not have a 𝑦-value of zero. So we’re not going to work out the remaining. And we can deduce that the option is (E).

But let’s double-check by substituting 𝑥 equals negative five, zero, and five in. When 𝑥 is negative five, 𝑦 is a fifth times negative five plus one. Well, a fifth times negative five is negative one. And when we add one, we get zero. Next, we let 𝑥 equal zero. So 𝑦 is a fifth times zero plus one, which is equal to one. Finally, we substitute 𝑥 equals five into this equation. And we get 𝑦 equals a fifth times five plus one. A fifth times five is one. So we’re doing one add one, which is two. And that’s the third value in our table.

We now notice that we have coordinates negative five, zero; zero, one; and five, two as we expected. And so the answer is option (E). The function represented by the graph shown has equation 𝑦 equals one-fifth 𝑥 plus one.