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I read a question and its explanation, and immediately had a doubt. the question was: $y = ax^3 + bx^2 + cx + 5$ is a curve which goes through $A(-2,0)$..blah blah..

d question was to find values of a,b,c

in the explanation, they took out the derivative of equation:

$$y' = 3ax^2 + 2bx + c$$

and in d above equation, they substituted x by -2 & equated it to 0, so it became:

$$0 = 12a -4b + c$$

can this be done? it was as if they treated y' as a function in x separately and assumed that it too passed through the point $(-2,0)$

another way of looking at it is that they figured out that dy/dx at $x= -2$ will be 0. (But I don't think this is correct)

so my question is, if a curve passes through some point, will its derivative also pass through the same point?

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    What is the rest of the question (..blah blah)? That might help answer your question.2011-08-04
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    I found this question online: The curve $y=ax^3+bx^2+cx+5$ touches the $x$-axis at $P(-2,0)$ and cuts the $y$-axis at the point $Q$ where its gradient is $3$. Find the equation of the curve completely. Would that be the question you read? If so, the word "touches" is significantly different than "goes through".2011-08-04
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    yes that is exactly the question. Can you explain how they just equated the derivative of y to 0 after putting x=-2 ?2011-08-04
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    @Rushil: right, but do avoid the profanities next time... :)2011-08-04

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