This sort of problem doesn't have one right answer. All that is required is that your answer be consistent: the line and the equation should match, and should be "reasonably close" to the data.
Your graph is a bit funny in that most vertical squares are marked as representing 10 units, but the first one is marked as 5. That's OK if you consider zero to be in the middle of the first square the way multiples of 10 are in the middle of squares above that. Most folks would expect the squares to be labeled 0 (at the bottom), then 10, 20, 30, and so on.
If you look at the first two data points, they are proportional with a slope of 15/2 = 30/4. However, the next two data points are not. It looks like you have drawn your line (and it should be a straight line—as with a ruler) through the first and last data points. If so, you can use those two points to find the equation.
The slope between them is ...
... (difference of y-values)/(difference of x-values) = (65 -15)/(8 -2) = 50/6 = 25/3
Then, the point-slope form of the equation of a line can be used. For slope m, the line through point (h, k) can be written as
... y = m(x -h) +k
If you use the first point and the slope calculated above, this becomes
... y = (25/3)(x -2) +15
... y = (25/3)x -(5/3) . . . . simplified
where y is the breaking weight in pennies and x is the number of layers.
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You will note that this line has a negative y-intercept. That is, a bridge with 0 layers is broken already. Of course, a bridge with 0 layers does not exist, so the domain of your equation would be for 2–8 layers (or so—covering the data you have, and maybe allowing for a little extrapolation).
Another way to draw the line is to give it a slope that matches the slope between the first two points and the last two points (y-difference of 15 for x-difference of 2), but has a slight positive offset to better match the last two points at the expense of matching the first two points. Such a line might have the equation
... y = (15/2)x +(5/2)
