# Lesson Video: Slope and Rate of Change Mathematics • 8th Grade

In this video, we will learn how to interpret the slope of a straight line as the rate of change of two quantities.

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### 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 ππ₯ plus π, or sometimes π¦ equals ππ₯ 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 ππ₯ 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 ππ₯ 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, weβll 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 weβll 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. Letβs 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 didnβt 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 ππ₯ 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.