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Topic 1 – part 2 Ratios and Proportional Relationships

Topic 1 – part 2 Ratios and Proportional Relationships

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Page 1: Topic 1 – part 2 Ratios and Proportional Relationships

Topic 1 – part 2Ratios and Proportional Relationships

Page 2: Topic 1 – part 2 Ratios and Proportional Relationships

Lesson 7.1.7Constant Rate of Change

•A rate of change is a rate that describes how one quantity changes in relation to another.

•A constant rate of change is the rate of change of a linear relationship.

Page 3: Topic 1 – part 2 Ratios and Proportional Relationships

Example 1Find the constant rate of change for the table.

Students # of Textbooks

5 15

10 30

15 45

20 60

The change in the number of textbooks is 15. The change in the number of students is 5.

The number of textbooks increased by 15 for every 5 students.

Write as a unit rate.

So, the number of textbooks increases by 3 textbooks per student.

Page 4: Topic 1 – part 2 Ratios and Proportional Relationships

Example 2The graph represents the number of T-shirts sold at a band concert. Use the graph to find the constant rate of change in number per hour.

To find the rate of change, pick any two points on the line, such as (8, 25) and (10, 35). 

or 5 T-shirts per hour

Page 5: Topic 1 – part 2 Ratios and Proportional Relationships

Lesson 8.3.2Slope

• The slope m of a line passing through points (x1, y1) and (x2, y2) is the ratio of the difference in the y-coordinates to the corresponding difference in the x-coordinates.

• As an equation, the slope is given by

, where x1 ≠ y1

Page 6: Topic 1 – part 2 Ratios and Proportional Relationships

Example 1

Find the slope of the line that passes through A(-1,-1) and B(2,3)

Slope formula

m = (x1, y1) = (–1, –1),

(x2, y2) = (2, 3)

Simplify.

Check When going from left to right, the graph of the line slants upward. This is correct for a positive slope.

Page 7: Topic 1 – part 2 Ratios and Proportional Relationships

Example 2

Find the slope of the line that passes through C(1, 4) and D(3, –2).

Slope formula

m = (x1, y1) = (–1, 4),

(x2, y2) = (3, –2)

Simplify.

Check When going from left to right, the graph of the line slants downward. This is correct for a negative slope.

Page 8: Topic 1 – part 2 Ratios and Proportional Relationships

Lesson 7.1.9Direct Variation

•When two variable quantities have a constant ratio, their relationship is called a direct variation.

•The constant ratio is called the constant of proportionality.

Page 9: Topic 1 – part 2 Ratios and Proportional Relationships

Example 1The time it takes Lucia to pick pints of blackberries is shown in the graph. Determine the constant of proportionality.

Since the graph forms a line, the rate of change is constant.Use the graph to find the constant of proportionality.  or or  It takes 15 minutes for Lucia to pick 1 pint of blackberries.

Page 10: Topic 1 – part 2 Ratios and Proportional Relationships

Example 2There are 12 trading cards in a package. Make a table and graph to show the number of cards in 1, 2, 3, and 4 packages. Is there a constant rate? a direct variation?

Numbers of Packages

1 2 3 4

Number of Cards 12 24 36 48

Because there is a constant increase of 12 cards, there is a constant rate of change. The equation relating the variables is y = 12x, where y is the number of cards and x is the number of packages. This is a direct variation. The constant of proportionality is 12.

Page 11: Topic 1 – part 2 Ratios and Proportional Relationships

Lesson 8.3.3Equations in y = mx form

•When the ratio of two variable quantities is constant, their relationship is called a direct variation.

Page 12: Topic 1 – part 2 Ratios and Proportional Relationships

Example 1

The distance that a bicycle travels varies directly with the number of rotations that its tires make. Determine the distance that the bicycle travels for each rotation.

Since the graph of the data forms a line, the rate of change is constant. Use the graph to find the constant ratio.

or or or  

The bicycle travels 80 inches for each rotation of the tires.

Page 13: Topic 1 – part 2 Ratios and Proportional Relationships

Example 2The number of trading cards varies directly as the number of packages. If there are 84 cards in 7 packages, how many cards are in 12 packages?

Let x = the number of packages and y = the total number of cards.

y = mx Direct variation equation

84 = m(7) y = 84, x = 7

12 = m Simplify.

y = 12x Substitute for m = 12.

Use the equation to find y when x = 12.

y = 12x

y = 12(12) x = 12

y = 144 Multiply.

There are 144 cards in 12 packages.

Page 14: Topic 1 – part 2 Ratios and Proportional Relationships

Lesson 8.3.4Slope-Intercept Form

•Linear equations are often written in the form y = mx + b. This is called the slope-intercept form.

•When an equation is written in this form, m is the slope and b is the y-intercept.

Page 15: Topic 1 – part 2 Ratios and Proportional Relationships

Example 1State the slope and the y-intercept of the graph of y = x – 3.

y = x – 3 Write the original equation.

y = 1x + (–3) Write the equation in the form y = mx + b.

↑ ↑y = mx + b m = 1, b = –3

The slope of the graph is 1, and the y-intercept is –3.

Page 16: Topic 1 – part 2 Ratios and Proportional Relationships

You can use the slope intercept form of an equation to graph the equation.

Example 2

Graph y = 2x + 1 using the slope and y-intercept.

Step 1 Find the slope and y-intercept.

y = 2x + 1 slope = 2, y-intercept = 1

Step 2 Graph the y-intercept 1.

Step 3 Write the slope 2 as . Use it to locate a second point on the line.

m = ← change in y : up 2 units

← change in x : right 1 units

Step 4 Draw a line through the two points.