Parallel Tangents of Cubic Graph (1 Viewer)

[flyingKangaroo]

Hey guys, I was wondering if anyone could me with this...

I have the graph of a function f(x)=(x+1)(x+4)(x+6). I've found the tangent at x=-6, the equation of which is y=10x+60. I then need to algebraically find the equation of another tangent on the curve which is parallel to the first.

Since I'm graphing this on Autograph I've managed to find a point where the tangent is parallel to the first, but since this can be done without the computer program I'm really interested to know how to do it. So yeah, I know that the tangent I'm trying to find will have a http://www.physicsforums.com/library.php?do=view_item&itemid=11gradient of 10 but that's about it...

Any help would be great

flyingKangaroo

 

[lyounamu]

Hey guys, I was wondering if anyone could me with this...

I have the graph of a function f(x)=(x+1)(x+4)(x+6). I've found the tangent at x=-6, the equation of which is y=10x+60. I then need to algebraically find the equation of another tangent on the curve which is parallel to the first.

Since I'm graphing this on Autograph I've managed to find a point where the tangent is parallel to the first, but since this can be done without the computer program I'm really interested to know how to do it. So yeah, I know that the tangent I'm trying to find will have a gradient of 10 but that's about it...

Any help would be great

flyingKangaroo

f(x) = (x+1)(x+4)(x+6) = (x^2+5x+4)(x+6) = x^3+5x^2+4x+6x^2+30x+24
= x^3+11x^2 + 34x +24
f'(x) = 3x^2+22x + 34
For it to be parallel, f'(x) = 10

3x^2+22x+34 = 10
3x^2+22x + 24 = 0
x = (-22+SR(22^2-4 . 3 . 24))/6 = -4/3 OR -6

So another tanget at x=-4/3

 

[bored of sc]

Hey guys, I was wondering if anyone could me with this...

I have the graph of a function f(x)=(x+1)(x+4)(x+6). I've found the tangent at x=-6, the equation of which is y=10x+60. I then need to algebraically find the equation of another tangent on the curve which is parallel to the first.

Since the equation of the first one is y = 10x+60 the gradient of the parallel tangent will be the same, that is m = 10.

 

[lyounamu]

I hope what I wrote above makes sense.

I first expanded the equation given to make it easier to be differentiated.

Then, I differentiated it to get the derivative form.

For the lines to be paralle, m1=m2 (whcih means 1st gradient = 2nd gradient)

Then, I found values for x where the lines are parallel.

I hope that cleared some confusions up~

 

[flyingKangaroo]

x = (-22+SR(22^2-4 . 3 . 24))/6 = -4/3 OR -6

So another tanget at x=-4/3

Okay, I get everything, and here I understand that you're solving for x I just don't fully understand your notation...

so its
x=(-22 + square root (22 squared minus 4 x 3 x 24)) over 6?

Yeah sorry that just really confused me...

 

[flyingKangaroo]

Oh wait I just realised you just use the quadratic formula...

Thanks!