For a linear function, the differences between consecutive $y$-values should be about the same. For an exponential function, the ratios $\dfrac{y_{\text{next}}}{y_{\text{current}}}$ should be about the same. Try both with the table.

Use the value of $y$ when $x = 0$ for the initial multiplier $a$. Then, use one step (for example, from $x = 0$ to $x = 1$) to estimate the factor $b$ that $y$ is multiplied by each hour.

Once you know the initial value and the approximate factor per hour, match them to the form $y = a(b)^x$ in the answer choices. Be sure both $a$ and $b$ agree with the data.

In Desmos, add a table and enter the $x$-values (0, 1, 2, 3, 4) in the first column and the corresponding $y$-values (120, 96, 77, 62, 50) in the second column to create plotted data points.

In separate lines, type each option exactly as given: y = 120 - 17x, y = 120(0.80)^x, y = 120(0.75)^x, and y = 110(0.85)^x. Make sure all four graphs are turned on so you can compare them to the plotted points.

Look at which graph passes closest to all the table points across the shown $x$-values. The equation whose curve most closely aligns with the plotted data is the best model.

Look at how $y$ changes as $x$ increases.

Compute the differences between consecutive $y$-values:

• From $120$ to $96$: change is $-24$
• From $96$ to $77$: change is $-19$
• From $77$ to $62$: change is $-15$
• From $62$ to $50$: change is $-12$

These differences are not constant, so a linear model (like $y = 120 - 17x$) is not a good fit. This suggests an exponential model, where the ratio between consecutive values is roughly constant.

For an exponential model $y = a\boldsymbol{(b)^x}$, the ratio $\dfrac{y_{\text{next}}}{y_{\text{current}}}$ is approximately the same each step.

Compute the ratios:

• $\dfrac{96}{120} = 0.8$
• $\dfrac{77}{96} \approx 0.802$
• $\dfrac{62}{77} \approx 0.805$
• $\dfrac{50}{62} \approx 0.806$

These are all very close to $0.8$, so the data are well modeled by an exponential function with a decay factor (base) of about $0.8$ per hour.

In an exponential model $y = a(b)^x$, $a$ is the initial value when $x = 0$.

From the table, when $x = 0$, $y = 120$, so $a = 120$.

Together with the decay factor $b \approx 0.8$, the model should look like

Among the choices, this matches the option $y = 120(0.80)^x$, so that is the best model for the data.