After keeping only equations whose vertex is near $x=20$, compare their predicted $y$-value at $x=20$ to the plot’s peak (about $70$ in hundreds).

Near $x=10$ and $x=30$, the plot’s $y$-values are close to $0$. Evaluate the remaining equations at one of those $x$-values to see which is closest.

Enter the points from the scatterplot as a list in Desmos, for example:

Define each choice as a function, for example:

• $f_1(x)=-0.7x^2+28x-210$
• $f_2(x)=-0.7x^2+28x-190$
• $f_3(x)=-0.6x^2+24x-170$
• $f_4(x)=-0.7x^2+25.2x-156.8$

Create a table for $x$ values like $10,20,30$ and compare the outputs $f_1(x), f_2(x), f_3(x), f_4(x)$ to the plotted points (peak near $x=20$, and small values near $x=10$ and $x=30$).

The function that best matches the peak near $(20,71)$ and stays closest to the points near $x=10$ and $x=30$ is $y=-0.7x^2+28x-210$.

From the scatterplot, the data rise to a maximum near $x=20$ with $y\approx 70$ (hundreds), then fall. Also, near $x=10$ and $x=30$, the $y$-values are close to $0$.

So a good quadratic model should have:

• an axis of symmetry near $x=20$,
• a peak near $y\approx 70$,
• small values near $x=10$ and $x=30$.

For $y=ax^2+bx+c$, the $x$-coordinate of the vertex is $x=-\frac{b}{2a}$.

Compute $x=-\frac{b}{2a}$ for each choice:

• $y=-0.7x^2+28x-210$: $x=-\frac{28}{2(-0.7)}=20$
• $y=-0.7x^2+28x-190$: $x=-\frac{28}{2(-0.7)}=20$
• $y=-0.6x^2+24x-170$: $x=-\frac{24}{2(-0.6)}=20$
• $y=-0.7x^2+25.2x-156.8$: $x=-\frac{25.2}{2(-0.7)}=18$

The scatterplot’s peak is near $x=20$, so eliminate $y=-0.7x^2+25.2x-156.8$ (its vertex is at $x=18$).

At $x=20$, the scatterplot’s $y$-value is about $71$ (hundreds).

Evaluate the remaining choices at $x=20$:

• $y=-0.7x^2+28x-210$ gives $y=70$.
• $y=-0.7x^2+28x-190$ gives $y=90$ (too high).
• $y=-0.6x^2+24x-170$ gives $y=70$.

Eliminate $y=-0.7x^2+28x-190$ because it overestimates the peak.

Now compare the last two candidates using a temperature near an end of the range, such as $x=10$ (where the scatterplot shows $y$ close to $0$).

• For $y=-0.7x^2+28x-210$:

• For $y=-0.6x^2+24x-170$:

The first model predicts a value much closer to the scatterplot near $x=10$ (and similarly near $x=30$). Therefore, the best model is $y=-0.7x^2+28x-210$.