#Equating the derivatives of two functions implies equality of the functions up to the linear term.

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warm geyser
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I have been given information about the graph two curves, y = m(x-a) and y = x^3 + x, and I am asked to (as the image shows) show that m = 3b^2 - 1 given that the two curves touch at x = b. I have almost finished, but I can't seem to justify why the constant term of m = ... should be a -1.
I've tried substituting it into y = m(x-a) = (3b^2 - 1)(x-a) to verify that (b, b^3) is a point in this curve when the constant term is -1, but I don't see a way to get rid of the a term without getting a in terms of b - and that task is the next question, so I don't believe that's the correct solution.
Is my understanding of derivatives wrong?

unknown.png
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also, here is a graph of the functions i've been given to help with visualisation:

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in the depicted case, m > 0 and a <= -1, but that's arbitrary

warm geyser
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am i wrong to say that

fresh palmBOT
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cat cat >w< pspspspsmeowmeow🐈

$\frac{d}{db} (mb-am) = \frac{d}{db}(b^3 - b) \nRightarrow m = 3b^2 - 1 \textrm{,but instead } m = 3b^2 + C$
326605555802308608.png
warm geyser
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ah, totally ignore this

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i got it all backwards