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

Question

amistre64 · Answer

how do we find inflection points?

anonymous · Answer

second deriv

amistre64 · Answer

good, and whats our 2nd derivative?

anonymous · Answer

6x+6

amistre64 · Answer

and that is zero for what value of x?

anonymous · Answer

-6

amistre64 · Answer

-1 ... but yes

what is the value of the first derivative at x=-1 to determine slope at x=-1

anonymous · Answer

i don't know...

amistre64 · Answer

whats our 1st derivative?

anonymous · Answer

3x^2+6x

amistre64 · Answer

right, and when x=-1 we get?  -3 right?

anonymous · Answer

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.

amistre64 · Answer

it is good for this case since its a cubic ...

anonymous · Answer

Yes, but still you have show it right? Otherwise teacher will deduct some marks ?!

amistre64 · Answer

slope at x=-1 is -3; we need to know the value of y at -1 to determine a point for this:

amistre64 · Answer

i dunno what a teacher does .... :)

anonymous · Answer

please continue :)

anonymous · Answer

They are there to deduct marks :P

anonymous · Answer

there is a point of inflection at -1

amistre64 · Answer

what is the value of y when x=-1?

y = x^3 +3x^2 + 2, when x=-1

anonymous · Answer

4

amistre64 · Answer

4?

(-1,4)

so, the equation of the tangent at the inflection is putting all the parts together:

tanY = -3x+3(-1)+4

anonymous · Answer

thank youuuuuuuuuuuu!

amistre64 · Answer

youre welcome

anonymous · Answer

can you look at my other question as well?