Use the image to answer the question.

Question

A decreasing curve with an arrow on the end is plotted on a coordinate plane. The curve begins at the point left parenthesis 0 comma 700 right parenthesis.

The value of a cell phone decreases exponentially at a rate of 24% each year after its release date. If the initial value of the phone is $700, the situation can be modeled by the equation v(t)=700(0.76)^t
, where v is the value of the cell phone t years since its release date. Interpret the graph of the equation and identify the true statement.

(1 point)
Responses

As t increases, v increases slowly at first and then quickly.
As t increases, v increases slowly at first and then quickly.

As t increases, v decreases slowly at first and then quickly.
As t increases, v decreases slowly at first and then quickly.

As t increases, v decreases quickly at first and then slowly.
As t increases, v decreases quickly at first and then slowly.

As t increases, v increases quickly at first and then slowly.
As t increases, v increases quickly at first and then slowly.

Answer 1

To interpret the graph of the equation \( v(t) = 700(0.76)^t \), which models the value of a cell phone that decreases at a rate of 24% each year, we can analyze the behavior of the function as time \( t \) increases.

The function represents an exponential decay, where the initial value of the phone is $700, and the base of the exponent (0.76) indicates that the phone retains 76% of its value each year (thus losing 24%).

1. As \( t \) increases: The term \( (0.76)^t \) gets smaller because 0.76 is less than 1, meaning the value \( v(t) \) will decrease.
2. Behavior of the decrease: Exponential decay functions start off decreasing more quickly when the value is higher, and as the value gets smaller over time, the rate of decrease slows down.

From this understanding, we can conclude:

• The value decreases quickly at first when the initial value is $700, and then as time progresses and the phone’s value approaches zero, the decrease becomes slower.

Therefore, the true statement is:

As t increases, v decreases quickly at first and then slowly.