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

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The slope of the tangent to any point on a curve is given by the value of the first derivative of the curve at the point. For y=x^3 - 7x^2 + 14x – 8

y’ = 3x^2 – 14x + 14

At the point where x = 1.

y’ = 3*1 – 14*1 + 14

=> y’ = 3

Now the value of y at the point where x = 1 is y = 1^3 – 7*1^2 + 14*1 – 8 = 1 – 7 + 14 – 8 = 0.

The equation of the line with a slope equal to 3, passing through (1, 0) is given as

(y – 0)/(x – 1) = 3

=> y/ (x – 1) = 3

=> 3x – 3 = y

=> 3x – y – 3 = 0

The required equation of the line is 3x – y – 3 = 0.

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