Find an equation of the tangent line to the curve at the given point y = x³ - 2x + 2 at (2, 4)?

Solution:

y = x³ - 2x + 2, (2, 4) (Given)

Find the first derivative, calculate the slope and find the equation.

We should first know the slope of the curve at (2,4)

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

dy/dx = 3x² - 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 = 4

On further calculation

c + 20 = 4 

c = 16

The equation of the tangent is y = 10 x- 16

Therefore, the equation of the tangent line to the curve is y = 10x -16

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Find an equation of the tangent line to the curve at the given point y = x³ - 2x + 2 at (2, 4)?

Summary:

The equation of the tangent line to the curve at the given point. y = x³ - 2x + 2, (2, 4) is y = 10x -16.