Following on from OLDMAN's 'Trivial Q' (1 Viewer)

[ND]

ii) Find the line that is twice tangent to the curve y=4x^4+14x^3+6x-10.

note: this is not as trivial.

edit: "Following on" is probably misleading, you don't use the result from OLDMAN's Q for this.

 

[Archman]

umm.. havent checked the calculations carefully, but is it something like:
y=(493/32)x-(4961/256)

 

[ND]

I haven't actually done the question yet.

Anyway which method did you use?

There is a real nice method where you would make a substitution for x which would cancel the x^3 (looks like it would be (x-7/8) in this case), then complete the sqaure getting something in the form (x^2 + a)^2+bx+c, and by seeing that the (x^2+a)^2 part would be twice tangent to the x axis, ax+b would be the line (then of course you have to sub (x+7/8) or whatever it is back in. I saw this method a while ago somewhere...

 

[OLDMAN]

A good problem, but demands some algebra using the textbook way ie. twice tangent means two pairs of double roots a,a,b,b. I would be interested in an "elegant" shortcut if you can remember it.

 

[ND]

Ok i'll give it a try:

y=4x^4+14x^3+6x-10 ...[1]

Substitue x-7/8 for x:

y=4(x-7/8)^4+14(x-7/8)^3+6(x-7/8)-10
= 4(x^2-147/16)^2 + 247x/16 - 357811/1024 ...[2]

Now the x-axis is obviously twice tangent to 4(x^2-147/16)^2, so y = 247x/16 - 357811/1024 is tangent to the transformation [2]. Subbing x+7/8 for x:

y = 247*(x+7/8)/16 - 357811/1024
y = 247x/16 - 343979/1024

Hmmm not the same as Archman's, i probably made a silly mistake somewhere...