Find the equation of the tangent to the curve y = x^3 - 7x^2 + 14x - 8 at the point where x=1.

We'll determine the y coordinate of the tangency point,
that is:

y = 1^3 - 7*1^2 + 14*1 -
8

y = 1 - 7 + 14 - 8

y =
0

So, the tangency point has the coordinates
(1,0).

Now, the expression of the first derivative
represents the tangent line to the given curve.

y' = x^3 -
7x^2 + 14x - 8

y' = 3x^2 - 14x +
14

For x = 1 => y' = 3 - 14 +
14

y' = 3

The slope of the
tangent line is m = 3.

The equation of the
tangent line, whose slope is m = 3 and the point of tangency is (1,0),
is:

y - 0 = m(x -
1)

y = 3(x -
1)

y = 3x -
3