Write the equation of the tangent to the graph of the function f (x) = x2 + 2 at the point x0 = 1

The equation of the tangent to the graph of the function f (x) at the point x = x0 has the following form:

y = f ‘(x0) * (x – x0) + f (x0).

Find the derivative of the function f (x) = x² + 2:

f ‘(x) = (x² + 2)’ = 2x.

Find the value of the derivative of the function f (x) = x² + 2 at the point х0 = 1:

f ‘(1) = 2 * 1 = 2.

Find the value of the function f (x) = x² + 2 at the point х0 = 1:

f (1) = 1² + 2 = 1 + 2 = 3.

We compose the equation of the tangent to the graph of the function f (x) = x² + 2 at the point x0 = 1:

y = 2 * (x – 1) + 3.

Simplifying this equation, we get:

y = 2x – 2 + 3;

y = 2x + 1.

Answer: the equation of the tangent to the graph of the function f (x) = x² + 2 at the point x0 = 1: y = 2x + 1.