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| Section | Objectives |
|---|---|
| Topic 1: Exponential Functions | - Interpret inputs and outputs for exponential functions - Interpret concavity for exponential functions - Interpret rates of change for exponential functions |
| Topic 2: Polynomial Functions | - Interpret concavity for polynomial functions - Interpret inputs and outputs for polynomial functions - Interpret rates of change for polynomial functions |
| Topic 3: Validity of Models | - Determine validity - Examine utility - Determine fit |
| Topic 4: Linear Functions | - Interpret inputs and outputs of linear functions - Interpret rates of change for linear functions |
| Topic 5: Functions and Algebra of Functions | - Derive conclusions based on graphs - Derive conclusions based on data - Derive conclusions based on notation |
| Topic 6: Graphical Depictions | - Interpret rates of change for situations - Interpret inputs and outputs for situations - Interpret concavity for situations - Interpret asymptotes for situations - Interpret maximum and minimum for situations |
| Topic 7: Logistic Functions | - Interpret rates of change for logistic functions - Interpret inputs and outputs for logistic functions - Interpret asymptotes for logistic functions |
>> Valid Applied-Algebra Test Objectives <<
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NEW QUESTION # 111
The graph shows the number of trees sold by a nursery, where the number of days since the beginning of the month is along the horizontal axis and the number of trees sold per day is along the vertical axis. The greatest number of trees are sold when the temperature is over 70°F.
When was the temperature likely over 70°F?
Answer: C
Explanation:
The problem states that the greatest number of trees are sold when the temperature is over 70°F. Therefore, we need to identify the days on the graph when tree sales are highest. The graph shows two major peaks in the number of trees sold per day. These peaks occur around day 2 and day 10. Since higher tree sales are linked to temperatures over 70°F, those two days are the most likely days when the temperature exceeded 70°F. Day 6 appears near a low point, so it is not likely associated with high temperature based on the given relationship.
Day 12 also does not match the main peaks. Therefore, the correct answer is C.
NEW QUESTION # 112
A vehicle is traveling away from a town at a fixed rate. After 1 hours, the vehicle is 200 miles from the town.
After 4 hours, the vehicle is 395 miles from the town.
Which function represents the distance, d, between the vehicle and the town after thours?
Answer: B
Explanation:
Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:
d(t)=mt+b
where:
m= " rate of change "
and
b= " initial distance from the town "
We are given two points:
(1#200)
and
(4#395)
These mean:
After 1hour, the vehicle is 200miles away.
After 4hours, the vehicle is 395miles away.
First, find the rate of change:
m=(395-200)/(4-1)
m=195/3
m=65
So the vehicle is moving away from the town at a rate of:
65 " miles per hour "
Now the function has the form:
d(t)=65t+b
Use the point (1#200)to find b:
200=65(1)+b
200=65+b
b=135
Therefore, the function is:
d(t)=65t+135
Check using t=4:
d(4)=65(4)+135
d(4)=260+135
d(4)=395
NEW QUESTION # 113
A vehicle is travelling away from a rest stop. The function
D(t)=48t+23
represents the distance from the vehicle to the rest stop at time t, where t is measured in hours and D is measured in miles.
What is the value of D(1.7)?
Answer: A
Explanation:
The function is:
D(t)=48t+23
This is a linear function because it has the form:
D(t)=mt+b
where:
48
is the rate of change, and:
23
is the starting distance from the rest stop.
We need to find:
D(1.7)
That means substitute:
t=1.7
into the function.
D(1.7)=48(1.7)+23
First multiply:
48(1.7)=81.6
Now add 23:
D(1.7)=81.6+23
D(1.7)=104.6
So the vehicle is:
104.6 miles
from the rest stop after 1.7 hours.
NEW QUESTION # 114
Through a local marketplace, a person makes livestock feeders to sell. The graph shows the functions modeling the cost and revenue.
What is the minimum number of livestock feeders that must be sold in order for the revenue to cover the costs?
Answer: D
Explanation:
This question asks for the break-even point, which is the point where:
Revenue=Cost
On the graph:
The solid blue line represents revenue.
The dashed blue line represents cost.
The revenue line starts below the cost line, meaning the person is not yet covering costs. The revenue line eventually crosses the cost line at a little more than 4 units.
Since the question asks for the minimum number of livestock feeders that must be sold, we need a whole number of feeders.
The graph shows the intersection occurs between:
4 and 5
So selling 4 feeders is not enough, but selling 5 feeders is enough for revenue to cover the costs.
Therefore, the correct answer is:
C
NEW QUESTION # 115
The logistic function f(x), whose graph is shown, models the number of people who have created a website account, where xrepresents the number of days since the website started and f(x)represents the number of people who have created an account.
What is one range of values for which the graph is concave down?
Answer: D
Explanation:
For a logistic graph, concavity changes at the inflection point.
Before the inflection point, the graph is usually concave up. This means the function is increasing faster and faster.
After the inflection point, the graph is concave down. This means the function is still increasing, but it is increasing slower and slower.
From the graph, the curve changes concavity around:
x=8
After x=8, the graph begins to level off as it approaches its upper limit.
So one interval where the graph is concave down is:
(8#16)
Therefore, the correct answer is:
# ( " A " )
NEW QUESTION # 116
......
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