The graph shows a decreasing exponential function.
The horizontal axis represents:
" Time in months "
The vertical axis represents:
" New user accounts per day "
The graph starts near 7,000new user accounts per day and decreases over time. However, it does not decrease toward 0. Instead, the graph levels off near a horizontal value around:
y=2,500
This horizontal value is the horizontal asymptote.
A horizontal asymptote shows the long-term value the function approaches. Since the asymptote is above 0, the model predicts that the number of daily new user accounts approaches a positive value, not zero.
So the justified statement is:
" The number of daily new user accounts will not decrease to 0. "

The graph shows the queue size changing over time.
The horizontal axis represents:
" Time in minutes "
The vertical axis represents:
" Queue size "
The graph is a decreasing exponential curve. It starts near 750people and decreases quickly at first. Then the curve begins to flatten near:
y=100
A horizontal asymptote is the horizontal line that the graph approaches as time continues.
Here, the graph approaches the horizontal line:
y=100
This means that as time passes, the number of people in the virtual queue gets closer and closer to 100.
It does not mean the queue decreases to 0, because the graph levels off near 100, not near 0.
Therefore, the correct interpretation is:
" As time passes, the number of people in the queue approaches 100. "

We are told that this year's number of property sales is 6 less than last year's number of property sales.
Let:
x= " number of properties sold last year "
and
H(x)= " number of property sales this year "
Since this year's sales are 6 less than last year's sales, the function rule is:
H(x)=x-6
The question says that the number of properties sold last year was:
x=330
Substitute 330into the function:
H(330)=330-6
H(330)=324
So the notation that correctly represents the number of property sales this year is:
H(330)=324
This means: if 330 properties were sold last year, then 324 properties are expected to be sold this year.

The average rate of change tells us how much the output changes, on average, for each 1-unit increase in the input.
Here:
x= " hours after 7:00 a.m. "
and
y= " estimated wait time in minutes "
From the graph, point Ais approximately:
A=(5,88.3)
and point Bis approximately:
B=(9,65)
Now use the average rate of change formula:
" Average rate of change " =(y_2-y_1)/(x_2-x_1 )
Substitute the values:
(65-88.3)/(9-5)
=(-23.3)/4
=-5.825
Rounded to two decimal places:
-5.83
The negative sign means the estimated wait time is decreasing from point Ato point B.
So the average rate of change is approximately:
-5.83 " minutes per hour "

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