How do you find algebraically the equation of the tangent to y =x^3 + 2x^2 -5x +3 at x = 2?

Question

anonymous · Answer

Equation of the tangent :D
That should already scream "Derivative!!!"

But in all seriousness, what's the derivative of this function? ^.^

clarence · Answer

umm...well, i don't really know :p But i found out that the point of contact is (2,9), is that correct? :p

anonymous · Answer

Yes.  The point of contact is always at (x , f(x) )

clarence · Answer

So i've done this just faintly, but if the point of contact is (2,9), then dy/dx should be 15?

anonymous · Answer

yeah, so that's the slope of your tangent line :)
So now you have a point (of contact) at (2,9) and a slope m = 15
Can you write the equation of the line now? ^.^

clarence · Answer

So, is it, 15x-21?? :D

anonymous · Answer

Yeah, precisely :)
y = 15x - 21

clarence · Answer

WOO!! :D starting to get the hang of this derivatice thingy :) But now it says, "find where the tangent cuts the curve again :O How do you do that?

anonymous · Answer

Don't know what that means ^

clarence · Answer

First it says, find the equation of the tangent (which we did) but then it says, find where the tangent cuts the curve again, assuming that curve is the whole y =x^3 + 2x^2 -5x +3 ?

anonymous · Answer

I'm thinking it's where the tangent line intersects the curve?

clarence · Answer

how do you find that?

anonymous · Answer

Set 15x - 21 = x^3 + 2x^2 - 5x + 3

clarence · Answer

are we suppose to find what x is or?

anonymous · Answer

Yeah, unfortunately, yes.
That's IF I was right in assuming that where the tangent cuts the curve is where it intersects the curve...

clarence · Answer

well, you haven't been wrong all night yet so :)

clarence · Answer

but how do you do it? :O I don't know what to do :/

anonymous · Answer

Me neither :D

anonymous · Answer

Well, obviously, x = 2 is a solution.

clarence · Answer

Huh? :p

anonymous · Answer

Because both the tangent line and the curve are equal to 9, when x = 2
Remember?
Because (2,9) is the point of contact.

clarence · Answer

Yes, so... what now? :p

anonymous · Answer

Well, rearrange the equation so that 0 is on one side.

clarence · Answer

and then i have to factorise? :O

anonymous · Answer

Yes.  But don't worry, you already know that one of the factors would be (x-2)
so you divide that polynomial by x-2, you will get a quadratic function, which may be factored, or you may use the quadratic formula.

clarence · Answer

okay, cool thanks :)

anonymous · Answer

No problem :)