Yeh, but after that, how do you find out whether the voltage starts as a positive or a negative voltage?
It should be D then, according to that equation shouldn't it?
No, it wouldn't be D.
According to the equation, the graph would be the derivative of the original graph.
Look at the original graph.
You get a small +ve gradient at the start. Then you get a large -ve gradient. Then you get large +ve gradient (same size, different direction). Then a large -ve gradient again. Then you get a small +ve gradient.
D implies the size of the gradients are constant throughout. This clearly isn't the case with the original graph. You can eliminate D simply on this basis.
The different sizes of the gradients have to be taken into account.
Technically, the positive and negative sides of the graph aren't explicitly stated. This means you could consider any side of the graph to be positive or negative. This is one of the tricks that iTute has employed with option B.
If the top half of the graph was defined to be positive, and the bottom half negative - then all the graphs would be incorrect. I would still say that B is the most correct in the situation that it was defined. D changes the direction of the voltage more often than it actually does - while B replicates the changes in direction well.