Interpreting Graphs: Linear Relation and Equations Insights

Here are a few recommended readings before getting started with this lesson.

Variables
Table of Values
Linear Equations in Two Variables
Solution of an Equation
Coordinate Pair
Coordinate Plane
Line

Here are a few practice exercises before getting started with this lesson.

a Which of the following equations is a linear equation in two variables?

b Consider the linear equation in two variables

y = - 3 x + 5 .

Which of the following coordinate pairs is a solution to the equation?

c Take a look at the tables of values shown below.

Which of the tables represents the linear equation in two variables

y = 2 x + 4 ?

Output Units	Input Units	Rate of Change Units	Possible Interpretation
meters	seconds	$\frac{meters}{second}$	Car speed over a certain period of time
bacteria	hours	$\frac{bacteria}{hour}$	Growth rate of bacteria in an experiment
U.S. dollars	hours	$\frac{U . S . dollars}{hour}$	Worker's hourly wage

Output Units

Input Units

Rate of Change Units

Possible Interpretation

meters

seconds

\frac{meters}{second}

Car speed over a certain period of time

bacteria

hours

\frac{bacteria}{hour}

Growth rate of bacteria in an experiment

U.S. dollars

hours

\frac{U . S . dollars}{hour}

Worker's hourly wage

Time (minutes)	Distance (kilometers)
$0$	$0$
$10$	$15$
$20$	$30$
$30$	$45$
$40$	$60$
$50$	$75$
$60$	$90$

Time (minutes)

Distance (kilometers)

0

0

10

15

20

30

30

45

40

60

50

75

60

90

Rate of Change $= \frac{d _{2} - d _{1}}{t _{2} - t _{1}}$
Times	Substitute	Evaluate
$10 - 20$ min	$\frac{3 0 km - 1 5 km}{2 0 min - 1 0 min}$	$1.5 \frac{km}{min}$
$0 - 60$ min	$\frac{9 0 km - 0 km}{6 0 min - 0 min}$	$1.5 \frac{km}{min}$

Rate of Change

= \frac{d _{2} - d _{1}}{t _{2} - t _{1}}

Times

Substitute

Evaluate

10 - 20

min

\frac{3 0 km - 1 5 km}{2 0 min - 1 0 min}

1.5 \frac{km}{min}

0 - 60

min

\frac{9 0 km - 0 km}{6 0 min - 0 min}

1.5 \frac{km}{min}

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(0, 0), (2, 50)$	$\frac{5 0 - 0}{2 - 0}$	$25$
$(2, 50), (4, 100)$	$\frac{1 0 0 - 5 0}{4 - 2}$	$25$

Rate of Change

= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Points

Substitute

Evaluate

(0, 0), (2, 50)

\frac{5 0 - 0}{2 - 0}

25

(2, 50), (4, 100)

\frac{1 0 0 - 5 0}{4 - 2}

25

Linear Equation: $y = 2 x + 3$
$x$	Substitute	Point
$x = 1$	$y y = 50 (1) + 30 = 80$	$(1, 80)$
$x = 3$	$y y = 50 (3) + 30 = 180$	$(3, 180)$
$x = 4$	$y y = 50 (4) + 30 = 230$	$(4, 230)$
$x = 7$	$y y = 50 (7) + 30 = 380$	$(7, 380)$

Linear Equation:

y = 2 x + 3

x

Substitute

Point

x = 1

y y = 50 (1) + 30 = 80

(1, 80)

x = 3

y y = 50 (3) + 30 = 180

(3, 180)

x = 4

y y = 50 (4) + 30 = 230

(4, 230)

x = 7

y y = 50 (7) + 30 = 380

(7, 380)

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(1, 80), (3, 180)$	$\frac{1 8 0 - 8 0}{3 - 1}$	$50$
$(4, 230), (7, 380)$	$\frac{3 8 0 - 2 3 0}{7 - 4}$	$50$
$(1, 80), (7, 380)$	$\frac{3 8 0 - 8 0}{7 - 1}$	$50$

Rate of Change

= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Points

Substitute

Evaluate

(1, 80), (3, 180)

\frac{1 8 0 - 8 0}{3 - 1}

50

(4, 230), (7, 380)

\frac{3 8 0 - 2 3 0}{7 - 4}

50

(1, 80), (7, 380)

\frac{3 8 0 - 8 0}{7 - 1}

50

Rate of Change	Rewrite
$\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m$	$y_{2} - y_{1} = m (x_{2} - x_{1})$

Rate of Change

Rewrite

\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m

y_{2} - y_{1} = m (x_{2} - x_{1})

Rate of Change	Rewrite
$\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m$	$y_{2} - y_{1} = m (x_{2} - x_{1})$
$\frac{y _{3} - y _{2}}{x _{3} - x _{2}} = m$	$y_{3} - y_{2} = m (x_{3} - x_{2})$
$⋮$	$⋮$
$\frac{y _{n} - y _{n - 1}}{x _{n} - x _{n - 1}} = m$	$y_{n} - y_{n - 1} = m (x_{n} - x_{n - 1})$

Rate of Change

Rewrite

\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m

y_{2} - y_{1} = m (x_{2} - x_{1})

\frac{y _{3} - y _{2}}{x _{3} - x _{2}} = m

y_{3} - y_{2} = m (x_{3} - x_{2})

⋮

⋮

\frac{y _{n} - y _{n - 1}}{x _{n} - x _{n - 1}} = m

y_{n} - y_{n - 1} = m (x_{n} - x_{n - 1})

Time (seconds)	Distance (meters)
$0$	$0$
$3$	$18$
$6$	$72$
$9$	$162$
$12$	$288$
$15$	$450$
$18$	$648$
$21$	$882$

Time (seconds)

Distance (meters)

0

0

3

18

6

72

9

162

12

288

15

450

18

648

21

882

Find the rate of change represented in each table or graph.

One way to use a graph to find the rate of change of a line is to measure the change in x and y between two known points, then divide the vertical change Δ y by the horizontal change Δ x. Rate of change=Δ y/Δ x Let's measure the vertical and horizontal change, also known as the rise and run of the graph, between two points that fall on the graph.

As the graph travels from left to right, the rise, or change in y, is 4. Similarly, the run, or change in x, is 3. Rate of change=rise/run ⇔ m=4/3 The rate of change is 43.

We can could also calculate the rate of change by using the Slope Formula. m = y_2-y_1/x_2-x_1 We will substitute the given points into the formula to solve for the slope. Notice that slope is another term for rate of change.

To determine the rate of change in the given table of values, let's see if we can identify a pattern of change between consecutive rows.

As we already found in Part A, the rate of change is the change of y divided by the change of x between two points that fall on the graph of the function. Rate of change=Δ y/Δ x Examining the table, we can see that every time x increases by 2, y increases by 8. We can substitute these values into the equation to calculate the rate of change.

The rate of change is 4.

Determine whether the rate of change of the line that models each linear relationship is positive, negative, zero, or undefined.

The length of a boat ride is $15$ miles long on the fifth day and $15$ miles long on the tenth day.

A barista earns $$ 10$ for $1$ hour of work and $$ 30$ for $3$ hours.

A student scores a $25$ on a test when answering no questions incorrectly, and they score $20$ points when answering five questions incorrectly.

The length of the boat ride is the same on the fifth day as it is on the tenth day. Therefore, when we move from day 5 to day 10, the y-value of the linear function does not change. Therefore, the function must be a constant, which means it is a horizontal line.

Recall that the rate of change of a horizontal line is zero.

The barista earns $10 for 1 hour of work and $30 for 3 hours. Therefore, the more hours the barista works, the more money they earn. This means the line that models the money the barista earns slants upwards from left to right.

The rate of change is positive.

The student scores 25 points on the test if they answer no questions incorrectly and 20 if they answer 5 questions incorrectly. The more questions that are answered incorrectly, the lower the test score becomes. The line that models the student's test score slants downward from left to right.

This means the rate of change is negative.

Consider the following situation.

During a snow day, the depth of the snow reaches $30$ centimeters after $1$ hour and $60$ centimeters after $3$ hours.

Which is the dependent variable, depth or time?

Find the rate of change.

From the given report, the snow depth changes as time passes by. Note that it may stop snowing at any time, but this would not cause the time to stop. On the contrary, if the time stops, so does the snow. Therefore, the passage of time is independent of whether it snows or not.

Consequently, we can conclude that the depth of the snow is the dependent variable and the number of hours is the independent variable.

The rate of change r shows the relationship between two changing quantities. In the formula below, Δ (Delta) stands for change. r=Δ dependent variable/Δindependent variable We know that the snow has reached a depth of 30 centimeters after 1 hour and 60 centimeters after 3 hours. We can calculate the changes in the dependent and independent variable by subtracting these numbers. Δ dependent variable:& 60- 30=30 Δ independent variable:& 3- 1=2 Now we can calculate the rate of change.

The rate of change for the depth of the snow is 15 centimeters per hour.

In the following diagram, we see the average weight gain after birth for four different mammals. The four mammals represented are human, rhinoceros, gray seal, and mouse.

Pair each mammal with the appropriate graph.

Which mammal shows the fastest rate of change?

What is the approximate rate of change for the weight of a gray seal?

Notice that the rates of change of the graphs will not give us any information to pair the graphs with the correct mammals. However, we can use the y-intercept in order to determine the graph that corresponds to each mammal. Let's first estimate the y-intercepts of each graph.

Graph	y-intercept
A	≈ 33kg
B	≈ 14kg
C	≈ 3.5 kg
D	≈ 0kg

Of the four mammals, the mouse is the smallest, both as a newborn and when fully grown. Therefore, the mouse must be paired with Graph D, whose y-intercept is close to 0. By contrast, the largest mammal of the four represented is the rhinoceros. Therefore, we can pair Graph A with the rhinoceros.

Finally, it may not be obvious how much a newborn gray seal weighs. However, we do know that a human newborn definitely does not weigh 14 kilograms, or about 31 pounds. Therefore, Graph C must be paired with human, and, by the process of elimination, we can pair B with gray seal.

Let's summarize the pairs we made. Graph A& → Rhinoceros Graph B& → Gray seal Graph C& → Human Graph D& → Mouse

The rate of change is given by the slope of a line. The steeper a line is, the greater the rate of change. Examining the four graphs, we see that the graph that increases the fastest is Graph A, which corresponds to the rhinoceros. The rhinoceros shows the fastest rate of change.

From part A, we know that Graph B corresponds to the gray seal. To determine the rate of change of a gray seal's weight, we must identify two points that lie on the line and then calculate the line's slope. Examining the graph, we can find two such points.

By substituting these points into the slope formula, we can determine the rate of change. Notice that we cannot be sure that the points have these exact coordinates. Therefore, we will use the approximation sign ≈ when calculating the rate of change.

We estimate the slope to be 0.2, which means the baby seal gains roughly 0.2 kilograms of mass per day.

Maya's small business made a profit of

$ 440

at the end of January and

$ 900

at the end of March. What is the rate of change in the profit for this time period?

A rate of change shows the relationship between two changing quantities. In the context of this situation, those quantities are the difference in profit and the difference in months. Notice that in this case, profit is the dependent variable and time is the independent variable. With this information, we can write a variation on the standard formula for the rate of change. Rate of change=Δ profit/Δmonths January is the first month of the year and March is the third month of the year. Therefore, we can find the difference in months by subtracting these values. We can find the difference in profit between the two months by using a similar calculation. Δ profit:& 900-440=460 Δ months:& 3-1=2 Now we can substitute these values into the formula to find the rate of change.

As we can see, the rate of change is $230 per month. In context, this means that Maya makes an additional $230 in profit every month when compared to the previous month.

	17 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Interpreting Graphs of Linear Equations

Catch-Up and Review

Extra

Extra

Hint

Solution

Interpreting the Rate of Change

Finding the Rate of Change Between Different Times

Proof

Pairs of Points in a Line Have a Constant Rate of Change

Consecutive Points in a Data Set with Constant Rate of Change Lie on a Line

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Interpreting Graphs of Linear Equations

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