Mr Phua bought a new car. After $t$ months, its value $V$ is given by $V = 250000e^{-pt}$, where $p$ is a constant. The value of the car after 36 months is $180 000.

1. Substitute the given values into the equation

When $t = 36$, $V = 180000$.
$180000 = 250000e^{-36p}$

2. Solve for $p$ using logarithms

$e^{-36p} = \frac{180000}{250000}$
$e^{-36p} = 0.72$
$-36p = \ln(0.72)$
$p = \frac{\ln(0.72)}{-36}$
$p \approx 0.0091296$

3. Round to the required number of significant figures to show the result

$p \approx 0.00913 \text{ (shown)}$

1. Substitute the given values into the model equation

$100000 = 250000e^{-0.00913t}$

2. Isolate the exponential term

$e^{-0.00913t} = \frac{100000}{250000}$
$e^{-0.00913t} = 0.4$

3. Take the natural logarithm of both sides to solve for t

$-0.00913t = \ln(0.4)$
$t = \frac{\ln(0.4)}{-0.00913}$
$t \approx 100.36$

4. Round to the nearest month

$t \approx 100 \text{ months}$

1. Convert the time to months and substitute into the model to find the expected value.

5 years = 60 months.
$V = 250000e^{-0.00913(60)}$
$V = 144555.15$

2. Compare the expected value with the offer and make a conclusion.

Since the expected value of the car ($\approx \char36{}144555$) is greater than the offer of $\char36{}130000$, he should not take up the offer.

1. Determine the $V$-intercept

When $t = 0$, $V = 250000e^0 = 250000$. The graph starts at $(0, 250000)$.

2. Determine the shape and asymptote of the graph

Since $p > 0$, $e^{-pt}$ represents exponential decay. As $t \to \infty$, $V \to 0$. The graph is a smooth curve decreasing from $250000$ towards the $t$-axis, which acts as a horizontal asymptote.