The question in Hebrew asks to find the equation of the tangent line to the function $y = x^2 + 2x$ at the point where $x = 1$.

Let's break down the steps to solve this problem:

1. Find the derivative of the function $y = x^2 + 2x$, which represents the slope of the tangent line at any point $x$.

   $\frac{dy}{dx} = \frac{d}{dx}(x^2 + 2x) = 2x + 2$

2. Calculate the slope at $x = 1$:

   $m = 2(1) + 2 = 4$ So the slope of the tangent line at $x = 1$ is $4$.

3. Find the coordinates of the point on the curve at $x = 1$:

   $y = (1)^2 + 2(1) = 1 + 2 = 3$ The point of tangency is $(1, 3)$.

4. Use the point-slope form of the equation of a line, which is:

   $y - y_1 = m(x - x_1)$ where $(x_1, y_1)$ is the point of tangency $(1, 3)$ and $m$ is the slope. Substituting the values:

   $y - 3 = 4(x - 1)$

   Simplifying this equation:

   $y - 3 = 4x - 4$ $y = 4x - 1$

Thus, the equation of the tangent line is $y = 4x - 1$.

Would you like more details on any step?

Here are 5 related questions to expand on this topic:

1. How do you find the tangent line to a function at a different point?
2. What does the derivative tell us about the function's behavior?
3. How can you verify that the tangent line touches the curve at exactly one point?
4. How do we interpret the geometric meaning of a tangent line?
5. What changes if we want the normal (perpendicular) line instead of the tangent line?

Tip: The point-slope form of a line is a very handy tool when you know the slope and a point on the line, so keep it in mind for similar problems!