# 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.

## Answers and Replies

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
Science Advisor
Homework Helper
Gold Member
Dearly Missed
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?

Oh! Thank you!