The value of a new car is \12\%V(t)t1 \le t \le 6$?

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SAT Nonlinear Functions Question 76 | Free SAT Prep | Aniko

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Question 76·Easy·Nonlinear Functions

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The value of a new car is $25,000. Each year, the value of the car is estimated to decrease by $12\%$ of its value at the beginning of that year.

Which equation models the estimated value $V(t)$ , in dollars, of the car after $t$ years, where $1 \le t \le 6$ ?

For percent growth or decay questions on the SAT, immediately rewrite the situation in the standard exponential form $A(t) = A_0(1 + r)^t$ , where $A_0$ is the initial amount and $r$ is the rate as a decimal (negative for decrease, positive for increase). Here, convert 12% to 0.12, subtract from 1 to get the decay multiplier $1 - 0.12 = 0.88$ , and plug into $A_0(\text{multiplier})^t$ . Then scan the choices for the one with the correct starting value out front and the correct multiplier (less than 1 for decay) raised to the power $t$ .

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Think about what remains, not just what is lost

If the car loses 12% of its value each year, what percent of its value does it still have after each year?

Convert the percent that remains to a decimal

Once you know what percent of the value remains each year, write that percent as a decimal (for example, 88% becomes 0.88). This decimal is your yearly multiplier.

Decide how to model repeated yearly change

When the same percent change happens year after year, should you add/subtract the percent repeatedly, or should you multiply by a common factor raised to the power $t$ ?

Check the base and exponent positions

In an exponential model for this situation, which quantity should be the starting value (multiplied in front), and which number should be raised to the power $t$ ?

Desmos Guide

Enter each option as a function

In Desmos, type four separate functions, one for each option, such as:

$f(t) = 25000(0.12)^t$
$g(t) = 25000(0.88)^t$
$h(t) = 25000(1.12)^t$
$k(t) = 0.12(25000)^t$ Use the same variable $t$ as in the problem.

Check the starting value and direction of change

Look at each graph near $t = 0$ . The correct model should give a value of 25000 when $t = 0$ (the car’s value when new) and should decrease as $t$ increases, since the car loses value each year. Eliminate any equations whose graphs do not start at 25000 or that increase over time.

Verify the 12% decrease in one year

Create a table in Desmos for each function to see values at $t = 0$ and $t = 1$ . For the correct model, the value at $t = 1$ should be exactly 12% less than the value at $t = 0$ . Compare the two table entries for each function and keep the one that shows this 12% drop.

Step-by-step Explanation

Interpret the percent decrease

The car loses 12% of its value each year, so it keeps the remaining 88% of its value each year. That means each year you multiply the current value by 0.88 (since 88% written as a decimal is $0.88$ ).

Write expressions for the first few years

Start with the initial value of $25000$ dollars.

After 1 year: $V(1) = 25000 \times 0.88$ .
After 2 years: $V(2) = 25000 \times 0.88 \times 0.88 = 25000(0.88)^2$ .
After 3 years: $V(3) = 25000(0.88)^3$ . You can see that each year adds another factor of $0.88$ .

Recognize the exponential pattern

When the same percent change happens every year, the situation is modeled by an exponential function of the form $\text{(initial value)} \times (\text{multiplier})^t$ . Here, the initial value is 25000 and the yearly multiplier is 0.88, so the model must use those numbers in that form.

Write the general function and match the choice

Putting it all together, after $t$ years the value is $V(t) = 25000(0.88)^t$ . Comparing with the options, this corresponds to choice B.

Strategy

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation