# Find the equation of the tangent to the curve

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.

arildno
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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*

arildno said:
Do you agree that we must have y'(x*)=3?
Yes.
arildno said:
That is, the derivatives of y(x) must be equal at x=1 and x=x*
I think I get it. But I still don't know how to get the answer.

arildno
$$3=3x^{2}-14x+14$$
$$3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0$$