### Video Transcript

In this video, we will learn how to
graph linear functions. We will identify the slope and
π¦-intercept of a line and learn how to use a table of values to draw the graph of a
function.

Letβs start by thinking about what
a linear function is. The general form of a linear
function is π¦ equals ππ₯ plus π or it may also be seen as π¦ equals ππ₯ plus
π. A linear function has two variables
π₯ and π¦, and there is no higher power of π₯ than simply π₯ to the power of
one. In both forms of this linear
function, the letter π, which is the coefficient of π₯, represents the slope or
gradient of the line. The slope of a line indicates how
steep the line is. The constant, π or π, indicates
the π¦-intercept of the linear function. So letβs have a look at some drawn
linear functions.

If we start with one of the most
basic linear functions, the line π¦ equals π₯, we can see that the gradient, the
coefficient of π₯, would be one. And thereβs also no constant term
at the end of this function, which indicates that the π¦-intercept would be
zero. And we could see here that it
passes through the π¦-axis at the point zero, zero. If we compare this to the line π¦
equals π₯ plus two, we can see that the gradient remains the same at one. But this time, the π¦-intercept is
plus two, and we can see on the graph that it crosses the π¦-axis at the point
two.

So if we wanted to draw the graph
of the function π¦ equals π₯ minus one, what would that look like? Well, the slope of the line would
be the same as the other two lines. And this time, the π¦-intercept
would be negative one. So if we want to consider how we
find the π¦-intercept of a line, if we are given that the function written down, we
can identify it from the equation in the form π¦ equals ππ₯ plus π. And if weβre only given the drawn
function, then we can identify the π¦-intercept simply by looking at where it
crosses the π¦-axis.

Weβll now have a look at how
changing the slope of a function changes how it looks when we graph it. Starting with our reference line of
π¦ equals π₯, we can see that when we graph the function at π¦ equals two π₯, this
line is much steeper. In other words, the coefficient of
π₯, the slope, is greater than it was in π¦ equals π₯. For every one unit on the π₯-axis,
the line π¦ equals π₯ goes up one unit on the π¦-axis. If we compare this to the line π¦
equals two π₯ then for every one unit on the π₯-axis, then it goes up by two units
on the π¦-axis.

So if we wanted to graph the
function π¦ equals three π₯, then for every one unit on the π₯-axis, it would go up
three units on the π¦-axis, which would look like this. We can notice that since none of
these three graphs had any constant term, then that means that they all go through
zero on the π¦-axis. The π¦-intercept is zero. So what happens if we have a
negative gradient, for example, the function π¦ equals negative π₯? We can see that lines with a
negative gradient slope downwards from left to right.

Now, we can think about how we
identify the slope of a line. The slope of a line is often
referred to as the rise over the run, where the rise is the increase or decrease in
the π¦-values. And the run is the increase or
decrease in the π₯-values. We can also use the formula that,
between two coordinates π₯ one, π¦ one and π₯ two, π¦ two, the slope is calculated
by π¦ two minus π¦ one over π₯ two minus π₯ one.

So, if on our line we chose the
coordinates one, three and zero, one, we can identify either of these coordinates
with the π₯ one, π¦ one or π₯ two, π¦ two values and substitute them into the
equation. Here, we have our π¦ two value of
one minus our π¦ one value of three over our π₯ two value of zero subtract our π₯
one value of one. This will simplify to negative two
over negative one, which would mean that the slope of this line is two. If we wanted to use the rise over
the run, we would have a rise of two and a run of one, giving us two over one, which
is two.

We can now think about how we would
graph a linear function. The first method is by using a
table of values. In this method, we take a few key
values for π₯ and work out the corresponding π¦-values. So, for example, if we wanted to
graph π¦ equals three π₯ plus one, we could choose π₯-values of zero, one, and
two. And we would then substitute these
values into our function to work out the π¦-value. Here, we would have π¦ is equal to
three times zero plus one as our first π¦-value, which would mean that π¦ would be
equal to one. For our second π₯-value of one, we
would substitute in π₯ equals one into our function to give us three times one plus
one, which is four. And for our final value when π₯
equals two, we have π¦ is equal to three times two plus one, which is six plus one,
giving us seven.

So now, weβve found three
coordinates which lie on this line, zero, one; one, four; and two, seven. So what we need to do is draw our
axis, plot these three coordinates, and draw a line through them. The advantage of a table of values
is that it gives us coordinates for our line even if weβre not entirely sure what it
looks like before we begin, which brings us to the second way in which we can graph
a linear function. And thatβs by identifying the
features of the function when itβs in the form π¦ equals ππ₯ plus π.

For example, if we were drawing the
graph of π¦ equals three π₯ plus one, we would know that the slope must be three,
meaning that, for every one on the π₯-axis, it would go three up on the π¦-axis. And since the π¦-intercept is one,
it would cross the π¦-axis at zero, one. When using this method, we need to
be careful and remember that a positive slope means that the line slopes upwards
from left to right. And a negative slope means that the
line slopes downwards from left to right.

Weβre now going to put into
practice what weβve learned by looking at some example questions. In the first question, weβll look
more closely at how to fill in a table of values.

Let us consider the function π of
π₯ equals eight π₯ minus 11. Fill in the table. Identify the three points that lie
on the line π¦ equals eight π₯ minus 11.

So here, we have our function π of
π₯ equals eight π₯ minus 11. We need to find the three missing
π¦-values by substituting in to π of π₯. So, in order to find our π¦-value
when π₯ is negative one, we substitute π₯ equals negative one into our function. So we have eight times negative one
minus 11. And since eight times negative one
is negative eight, we have negative eight minus 11, giving us negative 19. And thatβs our first missing value
in the table. For our next missing value, we
substitute π₯ equals zero into the function, giving us eight times zero minus 11,
giving us zero minus 11 which is negative 11. And so we have our second value in
the table.

For the final value in the table,
we have when π₯ is one, eight times one minus 11, which is equal to eight minus 11,
giving us an answer of negative three. So the answer to the first part of
the question is the three values in the table are negative 19, negative 11, and
negative three. The purpose of creating a table of
values for a function is to give us pairs of coordinates. So the first pair of coordinates
will be negative one, negative 19. The second will be zero, negative
11. And the third pair will be one,
negative three.

Looking at the graph to identify
the coordinate negative one, negative 19, we read the π₯-coordinate first going to
negative one and then downwards on the π¦-axis to negative 19. And so we can identify that point
πΌ will lie on the line. For the coordinate zero, negative
11, we donβt move on the π₯-axis and we go down to negative 11. And so we can identify that point
π» would also be on the line of this function. For the third coordinate, we can
identify if we go across on the π₯-axis one and down negative three, then this will
be the point πΊ. So the three points are πΌ, π», and
πΊ. If we had been asked to graph this
function and weβd found these three coordinates, we would have plotted them. And then we can draw a smooth line
going through the three points.

In the next example, weβll explore
what to do whenever we have a function that isnβt given in the general form π¦
equals ππ₯ plus π.

Consider the equation three π¦
equals six π₯ plus three over two. Rearrange the equation into the
form π¦ equals ππ₯ plus π. What are the slope and π¦-intercept
of the equation? Use the slope and intercept to
identify the correct graph of the equation.

Letβs begin this question by having
a look at the first part. Here, weβre asked to rearrange our
equation into the form π¦ equals ππ₯ plus π. We can remember that this form, π¦
equals ππ₯ plus π or often π¦ equals ππ₯ plus π, is the general form which
allows us to identify key parts of a graph. We can see that our equation has a
π¦ and an π₯ in the same way that the general form has a π¦ and an π₯. The difference here is that we have
a three π¦ instead of just π¦. So weβll need to divide both sides
of our equation by three.

This means that, on the right-hand
side, weβll be dividing by two multiplied by three. And as two times three is six, we
have π¦ equals six π₯ plus three over six. In order to simplify this fraction,
we can consider the right-hand side as six π₯ over six plus three over six. This is equivalent to π¦ equals π₯
plus a half. And this means we have rearranged
the equation into the form π¦ equals ππ₯ plus π. For the second part, we can recall
that when we have π¦ equals ππ₯ plus π, then the coefficient of π₯, the letter π,
indicates the slope or gradient of an equation. The constant term π represents the
π¦-intercept.

So when we take our equation in
this form, thatβs π¦ equals π₯ plus a half, the slope will be represented by the
coefficient of π₯, which in this case would be one. The π¦-intercept will be positive
one-half, which we can write as a half. And so we have answered the second
question. We can now take a look at the graph
options for the third part of the question. Weβve established that the
π¦-intercept of our graph will be at a half. We can see on our first graph that
the π¦-intercept is a half since the graph crosses through half on the π¦-axis. So this may be a possible
answer.

On the second graph, it crosses the
π¦-axis at negative a half. And so we can rule out the second
graph. The third graph has a π¦-intercept
of one, so this does not fit. And the fourth has a π¦-intercept
at negative one. The final graph has a π¦-intercept
of negative a half. So this wouldnβt work either. This means that we have one answer
left, but it is worth checking if the slope is equal to one.

We can recall that, to find the
slope of a line between two coordinates π₯ one, π¦ one and π₯ two, π¦ two, we
calculate π¦ two minus π¦ one over π₯ two minus π₯ one. We can select any two coordinates
that lie on the line. Here we have zero, a half and one,
three over two. It doesnβt matter which one we
designate as π₯ one, π¦ one and which we designate as π₯ two, π¦ two. So to find the slope, we have
three-halves subtract a half over one minus zero. And since three-halves subtract a
half is one and one subtract zero is one, we have one over one, which means that the
slope is indeed equal to one. And so our answer is that it is the
first graph which represents the equation three π¦ equals six π₯ plus three over
two.

By making a table of values,
determine which of the following graphs represents the equation π¦ equals a half π₯
plus one.

In this question, weβre asked to
make a table of values. This can be a very useful tool to
help us draw the graphs of functions. Setting up a table of π₯- and
π¦-values will give us sets of coordinates that will lie on the graph. So which π₯-values do we choose
then? Well, we can see that all of our
graph options roughly go from π₯ is negative four up to π₯ is four. So choosing a few values within
this range would be a sensible option.

In order to find the π¦-value for
each π₯-value, we substitute into the equation π¦ equals a half π₯ plus one. So when π₯ is negative two, we have
π¦ equals a half times negative two plus one. And a half times negative two will
give us negative one. So we have π¦ equals negative one
plus one, which is zero. And so we found that when π₯ is
negative two, π¦ is equal to zero. When π₯ is equal to negative one,
we have π¦ equals a half times negative one plus one. And a half times negative one is
negative a half plus one. And so we found another pair of
values in the table.

We can continue to complete the
values in the table. When we have completed the table of
values, we will be able to identify coordinates which lie on the line. Taking the first set of coordinates
of negative two, zero, we can see that only graphs C and E have this coordinate on
the line. Identifying the second coordinate
of negative one, a half, on graph C, we can see that this will be negative one,
negative a half, meaning that we can rule out option C. We can confirm that graph E does go
through the coordinate negative one, a half. It also goes through the coordinate
zero, one; one, three over two; and two, two. Therefore, the graph given in E
represents the equation π¦ equals a half π₯ plus one.

In this example, we saw how we
could use a table of values to recognize an equation of a function. In the next example, weβll use the
features of the equation to recognize the graph.

Which of the following is the
function represented by the shown graph? Option A, π¦ equals three-halves π₯
plus two. Option B, π¦ equals two π₯ plus
two-thirds. Option C, π¦ equals two-thirds π₯
minus two. Option D, π₯ equals two-thirds π¦
plus two. Or option E, π¦ equals two-thirds
π₯ plus two.

Letβs recall that the general form
of a linear equation is π¦ equals ππ₯ plus π or π¦ equals ππ₯ plus π, where the
constant term of π or π represents the π¦-intercept of the function. The value of π will indicate the
slope or gradient of the line. So therefore, if we were to
calculate the slope and the π¦-intercept of this drawn function, we could work out
what the function is. We can recall that, between two
coordinates π₯ one, π¦ one and π₯ two, π¦ two, the slope is equal to π¦ two minus π¦
one over π₯ two minus π₯ one.

We can select any two coordinates
on the line for π₯ one, π¦ one and π₯ two, π¦ two. But often the easiest ones to pick
are those which have integer values. We can see here that zero, two and
three, four both lie on the line. It doesnβt matter which coordinate
we designate as π₯ one, π¦ one and which we designate as π₯ two, π¦ two. To find the slope, we substitute in
our π¦ two and π¦ one values to give us four minus two over the π₯ two minus π₯ one
values, which is three minus zero. Simplifying this, we have a slope
of two-thirds.

To find the π¦-intercept, we look
at the graph to see where it crosses the π¦-axis. And that happens when the π¦-value
is two. We can then fill in our values of
slope and π¦-intercept into the general equation. We found that the slope π is equal
to two-thirds and the π¦-intercept of π is equal to two. And so our answer is that given in
option E, π¦ equals two-thirds π₯ plus two.

We can now summarize what weβve
learned in this video. We learned that a linear function
can be expressed as π¦ equals ππ₯ plus π or π¦ equals ππ₯ plus π. The value of π represents the
slope or gradient. And the constant value of π or π
represents the π¦-intercept of the line. The slope of a line is equal to the
rise over the run. It can also be calculated by π¦ two
minus π¦ one over π₯ two minus π₯ one for the coordinates π₯ one, π¦ one and π₯ two,
π¦ two. We saw that we can graph a function
either by creating a table of values and plotting the coordinates or by inspecting
the equation to find the values of the slope and π¦-intercept. This second method is a little bit
more difficult, but it does get easier the more we practice graphing linear
functions.