First find how much the cost changed between 2017 and 2022, then divide by how many years passed. This gives you the slope of the linear function.

In a function $C(t)$ that models cost $t$ years after 2017, what does $C(0)$ represent? Use the information in the problem to find that value.

Once you know the slope and the value of $C(0)$, look for the equation that has that slope as the coefficient of $t$ and that starting value as the constant term.

In Desmos, type the two points that the model must pass through: (0, 47.5) for 2017 and (5, 62.75) for 2022. These represent $(t, C(t))$ at the start and end of the interval.

Type each option as a separate function, for example:

• A(t) = 3.05t + 47.5
• B(t) = 3.05t + 62.75
• C(t) = -3.05t + 47.5
• D(t) = 3.5t + 47.5

Desmos will graph four lines.

Look at the graph and see which of the four lines passes exactly through both points (0, 47.5) and (5, 62.75). That line’s equation is the correct model.

The variable $t$ is the number of years after 2017.

From 2017 to 2022:

So the time interval is $5$ years, meaning the model must go from $t=0$ (year 2017) to $t=5$ (year 2022).

The cost rises from $47.50$ dollars to $62.75$ dollars.

Change in cost:

So the total increase over the 5 years is $15.25$ dollars.

For a linear function, the slope is

Compute this:

So the slope (rate of change) is $3.05$ dollars per year. Any correct equation must have slope $3.05$.

At $t=0$ (the starting year, 2017), the cost is $47.50$ dollars.

In slope-intercept form $C(t) = mt + b$:

• $m$ is the slope (we found $m = 3.05$),
• $b$ is the y-intercept, the value when $t=0$.

So $b$ must be $47.5$ (the starting cost). Any correct equation must therefore pass through the point $(0, 47.5)$ and have $b=47.5$.

Using slope-intercept form $C(t) = mt + b$ with $m=3.05$ and $b=47.5$ gives

This matches answer choice A, so $C(t)=3.05t+47.5$ is the correct linear model.