An equation of the line tangent to the graph of y=x^3+3x^2+2 at its point of inflection is?

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An equation of the line tangent to the graph of y=x^3+3x^2+2 at its point of inflection is?

Mathematics
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how do we find inflection points?
second deriv
good, and whats our 2nd derivative?

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Other answers:

6x+6
and that is zero for what value of x?
-6
-1 ... but yes what is the value of the first derivative at x=-1 to determine slope at x=-1
i don't know...
whats our 1st derivative?
3x^2+6x
right, and when x=-1 we get? -3 right?
Wait, second derivative is zero is not sufficient for point of inflection. One also needs the lowest-order non-zero derivative to be of odd order.
it is good for this case since its a cubic ...
Yes, but still you have show it right? Otherwise teacher will deduct some marks ?!
slope at x=-1 is -3; we need to know the value of y at -1 to determine a point for this:
i dunno what a teacher does .... :)
please continue :)
They are there to deduct marks :P
there is a point of inflection at -1
what is the value of y when x=-1? y = x^3 +3x^2 + 2, when x=-1
4
4? (-1,4) so, the equation of the tangent at the inflection is putting all the parts together: tanY = -3x+3(-1)+4
thank youuuuuuuuuuuu!
youre welcome
can you look at my other question as well?

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