Big Ideas Math: Modeling Real Life, Grade 8
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4. Comparing Linear and Nonlinear Functions
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Exercise 16 Page 300

Practice makes perfect

We are given a table that shows the cost y in dollars of x pounds of sunflower seeds.

Pounds, x Cost, y
2 2.80
3  ?
4 5.60

Notice that the cost for four pounds of seeds is the double of the cost for two pounds. Then, we can find the value of one pound of sunflower seeds by dividing 2.80 by 2. Cost per pound=2.80/2= 1.40 dollars Now that we have the cost per pound, we can find the value of three pounds of sunflower seeds by multiplying 3 by 1.40 dollars. 3(1.40) = 4.20 dollars The missing y-value is 4.20. This is the value that makes the table show a linear function because the rate of change is constant.

Since the table represents a linear function, recall the equation that represents a linear function. y= mx+ b Here, m is the slope and b is the y-intercept. We can to use the given table to help us calculate these values.
Pounds, x Cost, y
2 2.80
3 4.20
4 5.60
Let's start by substituting the points (2,2.80) and (3,4.20) into the slope formula and calculating m.
m = y_2-y_1/x_2-x_1
m=4.20- 2.80/3- 2
â–Ľ
Simplify right-hand side
m=1.40/1
m=1.40
The slope is 1.40, which means that the cost per pound of sunflower is 1.40 dollars. Now, remember that the y-intercept is the y-value where the line crosses the y-axis. This occurs when x=0. The value x=0 represents 0 pounds of sunflower seeds, which means that the cost is 0. Then, the y-intercept is equal to 0. y= 1.40x+ 0 → y=1.40x
Recall the linear function we wrote in Part B.

y=1.40x The cost will always increase as we increase the pounds of sunflower seeds we want to buy. Because of this, the function does not have a maximum value.