The average trade-in value, y thousand dollars, of a particular make of used car depreciates with the age of the car, x years, according to the following table.

1. Set up and label the coordinate axes with appropriate scales

Horizontal axis: Label as $x$ (Age, years) with a scale covering $2$ to $8$.
Vertical axis: Label as $y$ (Trade-in value, thousand dollars) with a scale covering $0$ (or a broken axis starting from $\approx 25$) to $55$.

2. Plot the given data points

Plot the 7 coordinate points:
$(2, 53.9), (3, 49.9), (4, 44.8), (5, 38.0), (6, 34.6), (7, 32.5), \text{ and } (8, 29.8)$.

1. State whether a linear model is appropriate based on the scatter diagram.

From the scatter diagram (plotted in part (i)), the data points follow a curve rather than a linear trend. Thus, a linear model is not appropriate.

2. State whether a linear model is appropriate based on the context of the question.

In the context of the problem, a linear model would imply a constant rate of depreciation, meaning that for a sufficiently large age $x$, the trade-in value $y$ would become negative. Since a car's value cannot be negative, a linear model is inappropriate.

1. State the long-term mathematical behaviour of a quadratic model and relate it to the context.

A quadratic model $y = ax^2 + bx + c$ represents a parabola. For large values of $x$ (long-term), if $a > 0$, $y$ will eventually start to increase, which contradicts the real-world fact that an aging car's value generally depreciates and does not increase. If $a < 0$, $y$ will eventually plunge to take increasingly negative values, which is impossible since a car's value cannot drop below zero. Therefore, a quadratic model is inappropriate.

1. Find the equation of the regression line of $\ln y$ on $x$ using a graphing calculator.

By keying the values of $x$ and corresponding values of $\ln y$ into a graphing calculator, we obtain:
$a = -0.10335 \text{ (to 5 s.f.)}$
$b = 4.1962 \text{ (to 5 s.f.)}$
So the fitted model is:
$\ln y = -0.10335x + 4.1962$

2. Use the regression line to estimate $y$ when $x = 5.5$.

Substitute $x = 5.5$ into the regression equation:
$\ln y = -0.10335(5.5) + 4.1962$
$\ln y = 3.6278$
$y = e^{3.6278} \approx 37.631$

3. State the final predicted value in context.

Since $y$ represents the trade-in value in thousands of dollars, the predicted value is $\char36{}37\,600$ (or $37.6$ thousand dollars, correct to 3 s.f.).