How do I find the equation of the tangent line to the curve defined by y = x³ + 2x + 2 at the point (2, 6)?

To find the equation of the tangent line to the curve at the given point (2, 6), we need to follow these steps:

The function is given as:

y = x³ + 2x + 2

The derivative of this function, which represents the slope of the tangent line at any point, is calculated using the power rule:

dy/dx = 3x² + 2

Next, plug in x = 2 into the derivative to find the slope of the tangent line at that point:

dy/dx = 3(2)² + 2 = 3(4) + 2 = 12 + 2 = 14

So, the slope of the tangent line at the point (2, 6) is 14.

The point-slope form of the equation of a line is given by:

y - y₁ = m(x - x₁)

Where (x₁, y₁) is the point on the line, and m is the slope.

Plugging in our values: (x₁, y₁) = (2, 6) and m = 14, we get:

y - 6 = 14(x - 2)

Now, distribute and solve for y:

y - 6 = 14x - 28

y = 14x - 28 + 6

y = 14x - 22

Final Result: The equation of the tangent line to the curve at the point (2, 6) is:

y = 14x - 22