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

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.
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Answer 1

The situation described in the problem involves an exponential decay model for the value of a cell phone, which decreases by 24% each year. The function given is \( v(t) = 700(0.76)^t \), where \( v \) is the value of the phone at time \( t \) (in years).

When analyzing the graph of this function:

• At \( t = 0 \), the value of the phone is $700.
• As \( t \) increases, the value of the phone decreases because of the multiplication by a number less than 1 (0.76).

Interpretation of the Graph:

1. Initially, the value of the phone is relatively high.
2. With each passing year (as \( t \) increases), the value decreases, but the rate of decrease slows down over time.
3. This means that in the first few years, the value drops quickly, but as time goes on, the amount it drops each year becomes smaller.

True Statement:

• The correct interpretation of the graph is: "As \( t \) increases, \( v \) decreases quickly at first and then slowly."

Therefore, the answer is: As t increases, v decreases quickly at first and then slowly.