Marilee is creating a garden in her yard. She needs to put up a fence to keep the animals out of it. She has 40 yards of fencing she can erect. This situation is modeled using the equation A=−w2+20w

To determine a reasonable scale for the domain (width \( w \)) of the garden, we need to consider the context of the problem. The total fencing available is 40 yards, and since the perimeter of the rectangle is given by the equation \( P = 2l + 2w \), we can rearrange it to find that \( l + w = 20 \), where \( l \) is the length of the garden.

In terms of width \( w \), the possible values should range from 0 to a maximum width that would allow for a non-negative area. In this case, if \( w = 0 \), the area is also 0, and if \( w = 20 \), the length \( l \) would be 0, which is not practical (and results in an area of 0 as well).

So the reasonable width range to consider is from \( w = 0 \) to \( w = 20 \). The more options you have in your table, the more detailed your results can be. Given the options you provided (1, 2, 5, 10), a scale of 2 would allow for a good balance between detail and simplicity.

Thus, a reasonable scale for the domain would be 2. The table would have values for \( w \) such as 0, 2, 4, ..., 20.