Question Video: Graphing Linear Functions by Making Tables Mathematics • 6th Grade

Let us consider the function 𝑓(π‘₯) = 8π‘₯ βˆ’ 11. Fill in the table. Identify the three points that lie on the line 𝑦= 8π‘₯ βˆ’ 11.

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Video Transcript

Let us consider the function 𝑓 of π‘₯ is equal to eight π‘₯ minus 11. Fill in the table. Identify the three points that lie on the line 𝑦 is equal to eight π‘₯ minus 11.

There are two parts to this question. Firstly, we need to complete the table by filling in the values of 𝑦, or 𝑓 of π‘₯, for the function eight π‘₯ minus 11. In the top row of our table, we’re given π‘₯-values of negative one, zero, and one. And we can calculate the 𝑦-values in the second row of our table by substituting the corresponding π‘₯-values into our function. Let’s begin with negative one. 𝑓 of negative one is equal to eight multiplied by negative one minus 11. And this is equal to negative 19. The first missing number in our table is negative 19.

Repeating this for π‘₯ equals zero, we have 𝑓 of zero is equal to eight multiplied by zero minus 11, which is equal to negative 11. When π‘₯ is equal to zero, 𝑓 of π‘₯, or 𝑦, is equal to negative 11. Finally, we need to calculate the 𝑦-value when π‘₯ equals one. Eight multiplied by one minus 11 is equal to negative three. The three missing values in the table are negative 19, negative 11, and negative three.

The second part of our question asks us to identify the three points on the graph that lie on the line 𝑦 is equal to eight π‘₯ minus 11. The easiest way to do this is to consider the three pairs of values we found in the first part of this question. These are the ordered pairs, or coordinates, negative one, negative 19; zero, negative 11; and one, negative three. And we can find the points on the graph by going along the horizontal or π‘₯-axis and then up or down the vertical 𝑦-axis. Going along to negative one and then down to negative 19, we find the point 𝐼. In the same way, the point 𝐻 has coordinates zero, negative 11. And finally, 𝐺 has coordinates one, negative three. The three points that lie on the line 𝑦 is equal to eight π‘₯ minus 11 are 𝐼, 𝐻, and 𝐺.

Since our equation is written in the form 𝑦 is equal to π‘šπ‘₯ plus 𝑏, we know it is a linear equation. We will have a straight line with slope, or gradient, equal to negative eight and a 𝑦-intercept of negative 11. Drawing this straight line, we see that it does indeed pass through the points 𝐼, 𝐻, and 𝐺.

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