PLEASE HELP!! what is the equation of the line tangent to the graph of y=x^3+3x^2+2 at its point of inflection. Explanation would be greatly appreciated!!!

Question

zepdrix · Answer

So we're going to build a tangent line.
Exactly where are we going to build it? Ah great question!
At the point where this function switches concavity.

zepdrix · Answer

So we'll need to get a second derivative in order to find this inflection point,
the location where the function changes concavity.

zepdrix · Answer

So get to work you. Find that derivative.

anonymous · Answer

the first derivative is 3x^2+6x

anonymous · Answer

and the second is 6x+6

anonymous · Answer

when we set that to zero we get x=1 which is the point of inflection

zepdrix · Answer

I think it's x=-1, ya?

anonymous · Answer

YES! sorry

zepdrix · Answer

Oh oh, this is a sneaky question.
I forgot, we also need to get the `slope` of that tangent line.
So we'll need the first derivative for that, ya?

anonymous · Answer

yes, the first derivative is 3x^2+6x

zepdrix · Answer

We want the slope AT x=-1

y'(-1) = m

anonymous · Answer

i see, i got m=-3 @zepdrix

zepdrix · Answer

Let's use a different Y for our tangent line so we don't confuse it with the original function, which they should have labeled f(x) -_- *grumble grumble*

anonymous · Answer

:) i understand your frustration lol

zepdrix · Answer

Oh, so we'll also need a `point` that the tangent line passes through.
Well the tangent line `touches` the curve at x=-1,
so they share that point, ya?

We'll need to plug x=-1 into the `original function` to find an ordered pair that satisfies our tangent line.

anonymous · Answer

i found that y=5 when i plugged in -1 into the original equation

zepdrix · Answer

Mmm I think it's 4, lemme check again.

anonymous · Answer

i mean 4

anonymous · Answer

no yes it is four

anonymous · Answer

and we then plug in everything we found into the point slope formula

zepdrix · Answer

Let's try to summarize what we've done so far,
cause this one is going kinda deep.

~We're trying to create a tangent line to our curve, it has form Y=mx+b.
~We found the location of the tangent line, x=-1.
~We found the slope of the tangent line, y'(-1)=-3.
~We found a point that lies on the tangent line, (-1,4).

zepdrix · Answer

Yes, if you prefer point-slope form :) that will work out nicely

anonymous · Answer

and i got y=-3x+1

zepdrix · Answer

yay good job!