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

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**

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

We have to find the tangent to the curve y = x^3 - 7x^2 + 14x - 8 at the point where x=1

At x = 1, y = 1 - 7 + 14 - 8 = 0.

The slope of the tangent to the curve at x = 1 is the value of y' at x = 1

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

At x = 1 it is 3 - 14 + 14 = 3

The equation of the tangent is (y - 0)/(x -1) = 3

=> y = 3x - 3

=> 3x - y - 3 = 0

**The equation of the tangent is 3x - y - 3 = 0**

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