How do I find the equation of the tangent line to the curve defined by the function y = x^3 + 2x + 1 at the point (3, 22)?

To find the equation of the tangent line to the curve at a specific point, we need to follow a systematic approach. In this case, we are given the curve:

y = x³ + 2x + 1

and the point at which we want to find the tangent line: (3, 22).

Step 1: Differentiate the Function

The first step is to compute the derivative of the function, which will give us the slope of the tangent line at any point on the curve. The derivative of y with respect to x is:

y’ = d/dx (x³ + 2x + 1) = 3x² + 2

Step 2: Evaluate the Derivative at x = 3

Now, we evaluate the derivative at the point x = 3 to find the slope of the tangent line:

y'(3) = 3(3)² + 2 = 3(9) + 2 = 27 + 2 = 29

Step 3: Use the Point-Slope Form to Write the Equation

With the slope (m = 29) and the point (3, 22), we can use the point-slope form of the equation of a line:

y – y₁ = m(x – x₁)

Substituting the known values:

y – 22 = 29(x – 3)

Step 4: Simplify the Equation

Now, we can simplify this equation:

Distributing the slope:

y – 22 = 29x – 87

Adding 22 to both sides:

y = 29x – 65

Conclusion

The equation of the tangent line to the curve at the point (3, 22) is: