To find the equation of the tangent line to the curve at the specified point, we first need to determine the derivative of the function, since the derivative will give us the slope of the tangent line. The function given is:
y = x³ + 2x + 1
Next, we calculate the derivative:
y’ = 3x² + 2
Now, we will 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
Now we have the slope of the tangent line, which is 29 and our point is (3, 22).
Using the point-slope form of the equation of a line, which is:
y – y1 = m(x – x1)
where m is the slope and (x1, y1) is the point, we can plug in our values:
y – 22 = 29(x – 3)
Now, we can simplify this equation:
y – 22 = 29x – 87
y = 29x – 65
So, the equation of the tangent line to the curve at the point (3, 22) is:
y = 29x – 65