Let's write rules for the functions, J(t) and B(t), first and then we can compare their slopes and y-intercepts. Finally, we will determine their domains and ranges and compare those as well.
Rule for J(t)
We know that Jeff earns $32 per hour. Thus, in t hours, he earns
32t dollars.
He also charges a $25 service fee. Therefore, the rule for the function J(t), which represents the amount Jeff earns in t hours, can be written as:
J(t)=32t+25.
Rule for B(t)
We must determine the rule for B(t) by looking at the graph. Let's identify the coordinates of two points on the graph to find the slope first!
Now we can substitute (0,40) and (4,160) into the Slope Formula!
m = y_2 - y_1/x_2 - x_1
m = 160 - 40/4 - 0
m = 120/4
m = 30
The slope is 30. Next, we need the y-intercept. We can tell that the function intercepts the y-axis as y=40. Therefore, b=40 and we can write the final rule for B(t).
B(t)=30t+40
Comparison of slopes and y-intercepts
We now have rules for the two functions, J(t) and B(t).
J(t)&= 32t+ 25
B(t)&= 30t+ 40
We can see that the slope of J(t) is greater than the slope of B(t). However, the y-intercept of B(t) is greater than the y-intercept of J(t).
Domains
A domain is the set of all possible input values for a function. In our functions, these are the possible values of t. We know that Jeff had a job that lasted 5.5 hours. We can think about the maximum number of hours he can work as t, and the minimum number of hours as 0, which is if he does not work at all. Then, the domain of J(t) is:
0 ≤ t ≤ 5.5.
From the graph, we can see that the possible values of t for B(t) start from t=0 and end at t=4. Thus, the domain of B(t) is:
0 ≤ t ≤ 4.
Ranges
Since the domain of J(t) is 0≤ t ≤ 5.5, its range is the set of all real values from J(0) to J(5.5). Luckily, we know that J(0) is the y-intercept of J(t). From the function rule, and knowing about slope-intercept form, we can tell that it is 25. Let's find J(5.5) by substituting t=5.5 in J(t).
J(t)=32t+25
J( 5.5)=32 ( 5.5)+25
J(5.5)=176+25
J(5.5)=201
The range of J(t) is:
25 ≤ J(t) ≤ 201.
We can identify the range of B(t) by looking at the outputs of B(t) on the graph. The function starts at B(t)=40 and ends at B(t)=160. Therefore, its range is:
40 ≤ B(t) ≤ 160.
