### Video Transcript

Let us consider the function π of
π₯ is equal to eight π₯ minus 11. Fill in the table. Identify the three points that lie
on the line π¦ 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 π¦, or π of π₯, for the function eight π₯ minus
11. In the top row of our table, weβre
given π₯-values of negative one, zero, and one. And we can calculate the π¦-values
in the second row of our table by substituting the corresponding π₯-values into our
function. Letβs begin with negative one. π 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 π of zero is equal to eight multiplied by zero minus 11, which is equal to
negative 11. When π₯ is equal to zero, π of π₯,
or π¦, is equal to negative 11. Finally, we need to calculate the
π¦-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 π¦ 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
π¦-axis. Going along to negative one and
then down to negative 19, we find the point πΌ. In the same way, the point π» has
coordinates zero, negative 11. And finally, πΊ has coordinates
one, negative three. The three points that lie on the
line π¦ is equal to eight π₯ minus 11 are πΌ, π», and πΊ.

Since our equation is written in
the form π¦ is equal to ππ₯ plus π, we know it is a linear equation. We will have a straight line with
slope, or gradient, equal to negative eight and a π¦-intercept of negative 11. Drawing this straight line, we see
that it does indeed pass through the points πΌ, π», and πΊ.