Find the equation of the tangent to the curve

1. Mar 23, 2005

footprints

Find the equation of the tangent to the curve $$y=x^3-7x^2+14x-8$$ at the point where $$x = 1$$. $$\text{Answer: }y = 3x -3$$
Find the x-coordinate of the point at which the tangent is parallel to the tangent at $$x = 1$$.

I need help on the second part.

2. Mar 23, 2005

arildno

Well, let x* be the point you seek.
What do you know of x*?
Do you agree that we must have y'(x*)=3?
That is, the derivatives of y(x) must be equal at x=1 and x=x*

3. Mar 23, 2005

footprints

Yes.
I think I get it. But I still don't know how to get the answer.

4. Mar 23, 2005

arildno

Well, do you agree that what you need to solve is the equation (written with "x"):
$$3=3x^{2}-14x+14$$
This can be rewritten as:
$$3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0$$
What must then x* be?

5. Mar 23, 2005

footprints

Oh! Thank you!