### Video Transcript

In this video, we will learn how to
interpret the slope or gradient of a straight line as the rate of change of two
quantities. We will begin by looking at a
definition of the rate of change and recall what we mean by the slope or gradient of
a line.

The rate of change of quantity π¦
with respect to quantity π₯ is the rate of change in π¦ to the change in π₯. This can be written as a fraction,
the change in π¦ divided by the change in π₯. This is expressed as a change in π¦
per one unit in π₯.

A linear function has a constant
rate of change between π¦ and π₯. We know that the graph of a linear
function is a straight line. This can be written in the form π¦
equals ππ₯ plus π, or sometimes π¦ equals ππ₯ plus π. The value of π refers to the slope
or gradient of the line. If our graph slopes upwards from
left to right, this value will be positive, whereas if it slopes downwards from left
to right, it will be negative. The value of π represents the
π¦-intercept. This is the point at which the
straight line crosses the π¦-axis.

The inverse of our last statement
is also true. Any function that has a straight
line graph is a linear function. We can calculate the slope or rate
of change of any straight line graph if we know two coordinates. If we have two points π΄ and π΅
that lie on a straight line with coordinates π₯ one, π¦ one and π₯ two, π¦ two, then
the rate of change or slope is equal to π¦ one minus π¦ two over π₯ one minus π₯
two. This is sometimes referred to as
the rise over the run. Whilst we can choose any two points
on the straight line, it is helpful to choose those with integer values.

We will now look at some questions
that involve calculating the rate of change on straight line graphs.

Use the graph to find out how many
action figures each boy has.

As our graph has a straight line,
we know that our function is linear. This means that for each extra boy,
there will be the same number of extra action figures. As the number of boys is on the
π₯-axis and the number of action figures on the π¦-axis, we can go vertically
upwards from one until we hit our straight line and then horizontally across to the
π¦-axis. This tells us that when we have one
boy, we have three action figures. This suggests that the answer is
three. Each boy has three action
figures.

We can check this by selecting some
other points on the graph. Reading off the graph, we see that
two boys have six action figures. We can repeat this for three, four,
five, and six boys which is equal to 18 action figures.

As our graph starts at the point
zero, zero, we can calculate the number of action figures that each boy has by
dividing the number of action figures by the number of boys at any of these
points. Six divided by two is equal to
three, and 18 divided by six is also equal to three. This confirms that the correct
answer is three action figures.

What is the initial value and the
rate of change for the function represented by the given graph?

As our graph is a straight line, we
know that the function will be linear. This means that it can be written
in the form π¦ equals ππ₯ plus π. The value of π is the slope or
gradient of the line. The value of π is the
π¦-intercept, which is where the graph crosses the π¦-axis.

The initial value of a function is
its value when π₯ equals zero. This is the same as the
π¦-intercept. We can see from the graph that our
line crosses the π¦-axis at the point zero, eight. Therefore, the initial value is
eight.

The rate of change of any function
is defined by the change in π¦ over the change in π₯. This is equivalent to the slope or
gradient of the line and can be calculated using the formula π¦ one minus π¦ two
over π₯ one minus π₯ two, where two points on the line have coordinates π₯ one, π¦
one and π₯ two, π¦ two.

As we already have one point on the
line zero, eight, we need to select any other point that lies on the line. It is useful to select a point with
integer coordinates. In this case, we will select the
point one, two. Point π΄ has coordinates zero,
eight. And point π΅ has coordinates one,
two. Subtracting the π¦-coordinates
gives us eight minus two. Subtracting the π₯-coordinates in
the same order gives us zero minus one. This can be simplified to six
divided by negative one. Dividing a positive number by a
negative number gives a negative answer. Therefore, the rate of change is
equal to negative six.

As our straight line is sloping
downwards from left to right, we know that our slope or gradient will be
negative. Therefore, this answer of negative
six seems sensible. As the rate of change is equal to
negative six and the initial value is eight, the equation of this straight line is
π¦ equals negative six π₯ plus eight.

Which of the following has a
greater rate of change? Is it graph (a) or (b) π¦ equals
two π₯ minus nine?

As our graph is a straight line,
this means it is a linear function. It can therefore be written in the
form π¦ equals ππ₯ plus π. The value of π is the slope or
gradient of the line, and π is the π¦-intercept. As the graph crosses the π¦-axis at
the point with coordinates zero, three, it is clear that the π¦-intercept is
three.

We can calculate the slope or
gradient of the straight line using the formula π¦ one minus π¦ two divided by π₯
one minus π₯ two. This is also known as the change in
π¦ over the change in π₯ or the rise over the run. In order to calculate the slope, we
need to select two points that lie on the line. In this case, weβll choose point π΄
with coordinates one, five and point π΅ zero, three. It would be just as easy to select
any of the other three points that are labeled: negative one, one; negative two,
negative one; or negative three, negative three.

Subtracting the π¦-coordinates
gives us five minus three. Subtracting the π₯-coordinates in
the same order gives us one minus zero. This is equal to two divided by
one, which is equal to two. The equation of the straight line
graph is π¦ equals two π₯ plus three. We were asked to decide which of
the two functions had the greater rate of change. The rate of change is the same as
the slope. As both of the equations have a
slope or gradient of two, they will also have a rate of change of two. We can therefore conclude that the
two functions have the same rate of change.

We can go one step further here and
say that any two functions with the same rate of change or slope will be parallel
lines. This means that they will never
cross or intersect.

The last two questions that weβll
look at involve calculating the rate of change from a table.

Using the given table, determine
the rate of change in temperature.

The rate of change can be
calculated by dividing the change in π¦ by the change in π₯. In this question, the rate of
change in temperature could be calculated by dividing the change in temperature by
the change in time. The change in temperature between
each of our values in the table is constant. It is plus four degrees. The change in time is also
constant. We are adding one hour, from 1:00
PM to 2:00 PM, 2:00 to 3:00 PM, and then 3:00 to 4:00 PM.

As both of these changes are
constant, we know that our relationship will be linear. The change in temperature is equal
to four degrees per hour. This is because four divided by one
is equal to four. For each hour that passes, the
temperature increases by four degrees.

This could be calculated by
selecting any two pairs of values from the table. Letβs consider the two end
columns. The change in temperature here can
be calculated by subtracting 55 from 67. The change in time is four minus
one. This is equal to 12 divided by
three, which equals four and proves that our answer is correct.

Jennifer wants to burn an extra
2000 calories a week. She can choose between the
following exercise classes. Which class burns the most calories
per hour? How long would she have to spend in
that class to reach her weekly goal?

Class A lasts one and a quarter
hours and burns 600 calories. In class B, 375 calories are burned
in 45 minutes. Class C lasts one hour and burns
400 calories. In order to calculate which class
burns the most calories per hour, we need to work out the rate of change. As the three durations are
different, we can add an extra column to the table to calculate the number of
calories burned per hour.

The duration of class C is one
hour, and 400 calories are burned. Therefore, she would burn 400
calories per hour. The duration of class B is 45
minutes, which is equal to three-quarters of an hour. As three-quarters is equal to the
decimal 0.75, we could say that 45 minutes is equal to 0.75 of an hour. To calculate the rate of change or
number of calories burned per hour, we need to divide the total calories burned by
the time in hours. For class B, we need to divide 375
by 0.75. This can be done on the calculator,
giving us an answer of 500. In class B, Jennifer would burn
calories at a rate of 500 per hour.

There is an easier way of working
this out if we didnβt have a calculator. We know that in 45 minutes,
Jennifer burns 375 calories. We could therefore calculate the
number of calories burned in 15 minutes by dividing by three. 300 divided by three is 100, and 75
divided by three is 25. Therefore, Jennifer would burn 125
calories in 15 minutes. Adding 45 minutes and 15 minutes
gives us 60 minutes, and adding 375 to 125 calories gives us 500 calories. As 60 minutes is equal to one hour,
this proves that our answer for class B was correct. Jennifer would burn 500 calories
per hour.

We could use a similar method to
this for class A, starting with 600 calories in one and a quarter hours or 75
minutes. However, the quicker way to work
out the number of calories burned per hour for class A would be to divide 600 by one
and a quarter. This is the same as dividing 600 by
1.25. This is equal to 480. Therefore, Jennifer burns 480
calories per hour in class A. The largest of these three values
is 500. We can therefore conclude that
class B burns the most calories per hour.

The second part of the question
asks us how long Jennifer would have to spend in that class to reach a weekly goal
of burning 2000 calories. We can calculate the time she would
need to spend by dividing the total calories she wants to burn by the number of
calories burned per hour. We need to divide 2000 by 500. Both the numerator and denominator
are divisible by 100, leaving us with 20 divided by five. As 20 divided by five is equal to
four, Jennifer would need to spend four hours in class B to burn 2000 calories.

We will now summarize the key
points from this video. The rate of change of quantity π¦
with respect to quantity π₯ is equal to the change in π¦ over the change in π₯. A rate of change is expressed as a
change in π¦ per change of one unit in π₯, for example, per hour. A linear function has a constant
rate of change between π₯ and π¦. The graph of a linear function is a
straight line. And inversely, any straight line
graph is a linear function. This function can be written in the
form π¦ equals ππ₯ plus π. π is the slope or gradient of the
line. This is the same as the rate of
change of a function. And when given any two points on
the line, this can be calculated using the formula π¦ one minus π¦ two over π₯ one
minus π₯ two. Where the two points have
coordinates π₯ one, π¦ one and π₯ two, π¦ two. The value of π is the
π¦-intercept. And this is the point at which the
straight line crosses the π¦-axis. This constant will sometimes be
labeled π instead of π.

Two functions that have the same
rate of change or slope will correspond to two parallel lines. The greater the slope or rate of
change, the steeper the line. Finally, if our graph slopes
upwards from left to right, we know that the rate of change will be positive. Whereas if it slopes downwards from
left to right, the rate of change or gradient will be negative.