SL 2.5—Modelling functions

At $t=0$, the cyclist is at the starting position, so $D(0)=0$ km.

$D(0)=15+c(2^0)=15+c=0$

$c=0-15=-15$

With $c=-15$, the model becomes $D(t)=15-15(2^{-t})$. As $t$ increases, $2^{-t}\to 0$ and $D(t)\to 15$, meaning the cyclist approaches (but never passes) the 15 km checkpoint, which is physically reasonable.

We need the value of $c$ such that $D(0)=0$, since the cyclist starts at the origin (0 km from start).

If $c=15$, then $D(0)=15+15(1)=30\ne 0$. This would place the cyclist 30 km away at the start — eliminate.

If $c=0$, then $D(0)=15+0=15$, meaning the cyclist is already at the checkpoint at $t=0$. This contradicts the premise — eliminate.

If $c=-7.5$, then $D(0)=15-7.5=7.5\ne 0$. This places the cyclist halfway — eliminate.

Only $c=-15$ gives $D(0)=15+(-15)(1)=0$, confirming the cyclist starts at the origin.

The model $C(h)=45h+12$ is a linear function of the form $f(x)=mx+c$, where $m=45$ is the rate of change and $c=12$ is the constant term (y-intercept).

In a cost model, the constant term $c$ represents a fixed cost — a charge that applies regardless of the number of hours rented. Here, $c=12$ represents the \12.00$ harbour fee.

At $h=0$ (before any rental time), $C(0)=12$, confirming that \12.00$ is paid even before any hours are used. This is the fixed harbour fee.

We need to determine the real-world meaning of the constant $12$ in the cost model $C(h)=45h+12$.

The cost per hour is the coefficient of $h$, which is $45$, not $12$. Eliminate 'The cost per hour of renting the boat'.

The minimum number of hours ($h\ge 1$) is given in the domain restriction, not by the value $12$. Eliminate 'The minimum number of hours the boat can be rented'.

The total cost for one hour is $C(1)=45(1)+12=57$, not $12$. Eliminate this option.

The remaining option — 'The fixed harbour fee charged regardless of how long the boat is rented' — correctly identifies $12$ as the constant (y-intercept), a fixed charge independent of $h$.