Use the two points to compute the slope $m$ of the line: $m = \dfrac{V_2 - V_1}{t_2 - t_1}$. Be careful with the sign—does the volume increase or decrease?

Once you know the initial volume and the rate (slope), write an equation of the form $V = V_0 + mt$ and then look for the choice that matches your equation.

In a new expression line, type (960-1200)/(45-0) and press Enter. The output is the constant rate of change (slope) in liters per minute; note that it should be negative because the tank is draining.

Use the slope value from Desmos as $m$ in the linear model $V = 1200 + mt$. Then compare this model to each answer choice and select the one that has the same initial value (1200) and the same slope $m$ (including the correct sign).

Treat time $t$ (in minutes) as the input and volume $V$ (in liters) as the output.

• At the moment draining begins, $t=0$ and the volume is $V=1200$, so one point is $(0,1200)$.
• After 45 minutes, $t=45$ and the volume is $V=960$, so another point is $(45,960)$.

These two points lie on the line that represents $V$ as a function of $t$.

Use the slope formula $m = \dfrac{\Delta V}{\Delta t}$ with the two points $(0,1200)$ and $(45,960)$:

So the tank is losing $\dfrac{16}{3}$ liters of water per minute (the negative sign shows decrease).

For a constant-rate situation, we can write a linear model in the form

where $V_0$ is the initial amount (the volume at $t=0$) and $m$ is the rate of change. Here:

• The initial volume is $V_0 = 1200$.
• The rate of change is $m = -\dfrac{16}{3}$.

Substitute these into the model to get the specific equation.

Substituting $V_0 = 1200$ and $m = -\dfrac{16}{3}$ into $V = V_0 + mt$ gives

Comparing with the answer choices, this matches choice D: $V = 1200 - \dfrac{16}{3}t$. This is the correct equation for the volume of water remaining in the tank as a function of time.