The graph represents a logistic growth model for viral infections. From day 9 to day 12, the curve is moving upward, so the number of infected people is increasing. During this interval, the graph is also becoming steeper, which means the rate of increase is getting larger. When a function is increasing and its slope is increasing, the correct description is that the quantity increases faster and faster. This interval occurs before the logistic graph begins to level off near its carrying capacity. The choices involving decrease are incorrect because the graph is not going downward. The choice "increases slower and slower" would apply to a later interval where the graph flattens. Therefore, the correct answer is A.

The value r
2
=0.99 indicates a very strong regression fit because it is close to 1. A strong fit means the model explains most of the variation in the data, so limited extrapolation can be appropriate. However, extrapolation should stay close to the observed data range. From the scatterplot, the data values appear to run from about x=5 to x=15. A reasonable extrapolation range extends slightly beyond those observed values, not too far outside them. The interval x=2.5 to x=17.5 stays close to the original data and is more appropriate than the wider interval x=0 to x=20. Therefore, the correct answer is B.

The graph shows a logistic growth function for conference registration. From week 1 to week 7, the number of registrants is increasing because the graph rises as time moves forward. The curve also becomes steeper throughout this interval, meaning the number of registrants is increasing at a faster rate each week. In Applied Algebra, when a graph rises and its slope increases, the correct interpretation is "increases faster and faster." The graph is not decreasing, so both decreasing options are incorrect. The graph is also not flattening during this interval, so "increases slower and slower" does not match. Therefore, the correct answer is C.

The graph shows the temperature of an object changing over time.
The horizontal axis represents:
" Time in minutes "
The vertical axis represents:
" Temperature in degrees Fahrenheit "
The curve is decreasing quickly at first and then begins to level off. This is the shape of an exponential decay function.
A horizontal asymptote is a horizontal line that the graph approaches as time increases.
Because a horizontal asymptote is a horizontal line, its equation must have the form:
y= " constant "
From the graph, the temperature approaches about:
25^ # F
So the horizontal asymptote is:
y=25
This means the object's temperature gets closer and closer to 25^ # Fover time.

This problem uses an exponential growth model. The initial portfolio value is 3100, and the growth factor is
1.059. Since the growth factor is greater than 1, the portfolio increases over time. To find the value after 10 years, substitute x=10 into the function: g(10)=3100(1.059)
10
First evaluate the exponent, then multiply by 3100. This gives g(10)#5499.4758. Rounded to the nearest cent, the value is $5,499.48. The answer must be written as money because the function models the value of an investment portfolio. Among the choices, only option B matches the correctly evaluated exponential expression. Therefore, the correct answer is B.