# 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.**

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**

To find the tangent at x=1 to the curve y = x^3 - 7x^2 + 14x - 8.

The tangent to a curve y = f(x) at x1 is y-y1 = f'(x1)*(x-x1), where y1 = f(x1).

y = f(x) = y = x^3 - 7x^2 + 14x - 8.

dy/dx = f'(x) = y = (x^3 - 7x^2 + 14x - 8)'

f'(x) = 3x^2-14x+14.

f'(x1) = f'(1) = (3*1^2-14*1+14) = 3.

y1 = f(1) = 1^3-7*1^2+14*1-8 = 0

Therefore the tangent at (x1,y1) = (1,0) is given by:

y-0 = 3(x-1).

y -3x +3 = 0.