# Tangent.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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For the equation of the tangent we need the slope of the tangent and one point it passes through. The slope of a tangent to any curve at a point is the value of the first derivative at that point.

For the curve y = x^3 - 7x^2 + 14x - 8

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

at x = 1, y' = 3 - 14 + 14 = 3

Also, if x = 1, y = 1 - 7 + 14 - 8 = 0

The tangent has a slope 3 and passes through (1, 0)

This gives the equation of the tangent as (y - 0)/(x -1 ) = 3

=> 3x - y - 3 = 0

**The required equation of the tangent 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**