If a point is far to the right and below the trend, would it make the best-fit line flatter or steeper?

Use a small $x$ (like $x=1$) and a larger $x$ (like $x=8$) to see which line stays closest to the remaining points.

In Desmos, create a table and enter the 8 points that are not labeled P (use the x-values as the first column and y-values as the second column).

In separate expressions, enter the four equations from the answer choices (for example, y=3.2x+54, y=2.4x+56, and so on).

Visually compare which line runs through the middle of the 8 plotted points (with about as many points above as below). The line that best matches this is the one you should select.

From the graph, the points from about $x=1$ to $x=8$ follow an upward trend, but point P is at a large $x$-value and has a much smaller $y$-value than nearby points.

Because P is to the right of the other points and below the trend, it pulls the right side of the line downward.

Removing P would therefore make the new line of best fit steeper (it would have a larger slope than $2.4$).

A line that fits the remaining cluster should still predict a score in the high 50s when $x$ is around $1$, and a score near $80$ when $x$ is around $8$.

Among the choices, $y=3.2x+54$ has a larger slope than $2.4$ and predicts

• at $x=1$: $y=3.2(1)+54=57.2$ (close to the left-side points)
• at $x=8$: $y=3.2(8)+54=79.6$ (close to the upper-right points)

So the most likely new line of best fit is $y=3.2x+54$.