How can the common tangent of two curves be found without using derivatives?

[cruisx]

Homework Statement

Hi guys, need some help finding the common tangent of two curves.
Two curves y = 3X³+6X²+6X+3 and y = 3X³+6X+3 touch each other. Find the common tangent.

Homework Equations

y = 3X³+6X²+6X+3 and y = 3X³+6X+3

The Attempt at a Solution

well i made the two equations equal each other and then i found the X values. After that i subbed the values in the the original equations to get my y values. My points ended up being
(0,3) and (-1,0)

is the correct? what do i do after this?
First time with calculus so not sure what to do.

 

[tiny-tim]

Hi cruisx!

y = 3X³+6X²+6X+3 and y = 3X³+6X+3
…
My points ended up being
(0,3) and (-1,0)

(0,3) yes, but how did you get (-1,0) ?

anyway, now find the tangent at (0,3).

 

[cruisx]

Hi cruisx!


(0,3) yes, but how did you get (-1,0) ?

anyway, now find the tangent at (0,3).

well...um i forget how i got (-1,0) so i only needed to find (0,3)? and then after i find the tangent i am done? I think i got -1,0 because my x values were -1 and 0 so i subbed -1 into one of my equations to get (-1,0). So i should use (0,3) to find the common tangent?

 

[tiny-tim]

I think i got -1,0 because my x values were -1 and 0 so i subbed -1 into one of my equations to get (-1,0).

Yeah, I guessed that … but how did you get x = -1?

(I got 6X² = 0)

So i should use (0,3) to find the common tangent?

Yes.

 

[cruisx]

Hi, so it seems that our next unit is derivatives so i am guessing i will have to find the tangent differently? can someone tell me how to find the tangent using (0,3) because it seems that it has to lead to a derivative or what ever that is. Will find out tomorrow.

 

[tiny-tim]

Hi cruisx!

(just got up :zzz: …)

Hi, so it seems that our next unit is derivatives so i am guessing i will have to find the tangent differently? can someone tell me how to find the tangent using (0,3) because it seems that it has to lead to a derivative or what ever that is. Will find out tomorrow.

hmm … without derivatives?

ok … try this …

you know the tangent (to either curve) is a straight line through (0,3),

so it has to be y = Ax + 3, for some value of A.

now try a tiny-teeny-weeny value of x (ie very close to 0) in the equation for the curve, and see what for value of A it looks most like y = Ax + 3.

(if you prefer, change the cooridnate to y' = y - 3, so the curves meet at the origin)