# Tangent line

1. Oct 25, 2006

### k3232x

Hi, im having problems with the following problem. The main issue is actually starting the problem.

Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line.

2. Oct 25, 2006

### cliowa

Say I'd give you a point p0=(x0,y0), where x0 and y0 are certain, fixed real values. How would you calculate the tangent line?

3. Oct 25, 2006

### Aspekt

You first need to find the derivative of y = x^4 - 2x^2 - x

4. Oct 26, 2006

### k3232x

I took the derivative of the equation and got:

y`= 4x^3 - 4x - 1 and put that into the equation of a tangent to the curve:

y=(4x^3 - 4x - 1)x + c ; where c is the intercept.

Then replaced y with the original equation:

x^4 - 2x^2 - x = (4x^3 -4x - 1)x + c

Then solved for the intercept:

c= -3x^4 + 2x^2

Need 2 points which have a common tangent, (a,b) and (c,d) then:

4a^3 -4a -1 = 4c^3 -4c -1 and

-3a^4 + 2a^2 = -3c^4 - 2c^2

Now here's where I am stuck. I don't know how to solve these equations which would then give me the points (a,b) and (c,d) that i need for a common tangent line.

5. Oct 26, 2006

### HallsofIvy

Try writing
$$4a^3 -4a -1 = 4c^3 -4c -1$$
and
$$-3a^4 + 2a^2 = -3c^4 - 2c^2$$
as
$$4(a^3- c^3)- 4(a-c)= 0$$
and
$$-3(a^4- c^4)+ 2(a^2- c^2)= 0$$
and factor:
$$a^4- c^4= (a- c)(a^3+ a^2b+ ab^2+ b^3)$$
$$a^3- c^3= (a- c)(a^2+ ab+ b^2)$$
$$a^3- b^2= (a- c)(a+ b)$$
Obviously a= c so a- c= 0 satisfies both of your original equations. What else will?

Last edited by a moderator: Oct 26, 2006
6. Oct 26, 2006

### GoldPheonix

Take the derivative of the f(x).

y' = 4x^3 - 4x - 1

That is your tangant line function, now we need to find only two points that have the same tangent line. Find the tangent line at a point. In this case, we'll find it for 1.

y' = 4(1)^3 - 4(1) - 1
y' = -1

Now, let's plug that back into the equation and try to find the other values that have the same slope.

-1 = 4x^3 - 4x - 1
0 = 4x^3 - 4x
0 = 4x(x^2 - 1)
0 = 4x(x-1)(x+1)

Which means that it will have a slope of -1 at (-1,-2); (1,-2); and (0,0).

Let's check that out.

f'(-1) = 4x^3 - 4x - 1 = -1
f'(0) = 4x^3 - 4x - 1 = -1
f'(1) = 4x^3 - 4x - 1 = -1 (From previous steps)

Confirmed.

Those points up above are points that have the same tangent line.

7. Oct 27, 2006

### HallsofIvy

No, that gives points that have parallel tangent lines, not necessarily the same tangent line.

8. Oct 28, 2006

### GoldPheonix

Oh, I see. They want the points where they have the same linear tangent line function. Well, in that case:

y + 2 = -1(x -1) => y = -x -1
y + 2 = -1(x + 1) => y = -x -3
y + 0 = -x

In which case, those points do not have equal tangent line functions. Yeah, I would use a different method to find things of equal tangent line functions.