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# Find an equation of the tangent line to the curve at the given point y = x^{3} - 2x + 2 at (2, 6)?

**Solution:**

y = x^{3} - 2x + 2 at (2, 6) (Given)

We should first know the slope of the curve at (2,6)

Slope at any point on the curve is written by its first derivative as

dy/dx = 3x^{2} - 2

Slope at x = 2

m = 3(2)^{2} - 2 = 12 - 2 = 10

Equation of a tangent is given as

c + mx = y

Where m is the slope

Substituting the values

c + 10 × 2 = 6

On further calculation

c + 20 = 6

c = 6 - 20 = -14

Therefore, the equation of the tangent line to the curve is y = 10x - 14.

## Find an equation of the tangent line to the curve at the given point y = x^{3} - 2x + 2 at (2, 6)?

**Summary:**

The equation of the tangent line to the curve at the given point y = x^{3} - 2x + 2 at (2, 6) is y = 10x - 14.

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