How do you find the slope of the tangent line to the function y = x² + 2x + 3 at the point where x = 1?

To find the slope of the tangent line to the function y = x² + 2x + 3 at the point x = 1, we first need to calculate the derivative of the function. The derivative represents the slope of the tangent line at any given point on the curve.

The first step is to differentiate the function:

y = x² + 2x + 3

Using basic differentiation rules:

y' = d/dx(x²) + d/dx(2x) + d/dx(3)

This results in:

y' = 2x + 2

Now, to find the slope at the specific point where x = 1, we substitute x with 1 in the derivative:

y'(1) = 2(1) + 2

This simplifies to:

y'(1) = 2 + 2 = 4

Therefore, the slope of the tangent line to the graph of the function y = x² + 2x + 3 at the point x = 1 is 4.