The problem is asking for the equation of the tangent line to the function $f(x) = x^2 + x$ at the point $(1, 2)$, and it provides the derivative $f'(x) = 2x + 1$.

Step-by-Step Solution:

1. Equation of a Tangent Line: The general form for the equation of the tangent line is: $y - y_1 = m(x - x_1)$ where $m$ is the slope of the tangent line, and $(x_1, y_1)$ is the point of tangency.

2. Slope of the Tangent Line: The slope of the tangent line at a given point is the derivative of the function at that point. The derivative is provided as $f'(x) = 2x + 1$.

   To find the slope at $x = 1$: $f'(1) = 2(1) + 1 = 3$ So, the slope $m = 3$.

3. Point of Tangency: The point of tangency is given as $(1, 2)$.

4. Substitute into the Tangent Line Equation: Using the point $(1, 2)$ and the slope $m = 3$ in the equation of the line: $y - 2 = 3(x - 1)$ Simplifying this: $y - 2 = 3x - 3$ $y = 3x - 1$

Final Answer:

The equation of the tangent line is $y = 3x - 1$.

Would you like more details on any step? Here are some related questions:

1. What is the geometric meaning of a tangent line?
2. How do you find the derivative of a function like $f(x) = x^2 + x$?
3. Can you explain how the point-slope form works in different situations?
4. How would you find the tangent line if the derivative wasn't given?
5. What happens when the slope at a point is zero?

Tip: Remember, the slope of the tangent line is equal to the value of the derivative at the point of interest.