The problem in the image asks us to find the equation of the tangent line to the curve $y = 2e^x$ at the point $(0, 0)$. Let us solve it step by step.

Solution:

1. Find the derivative of the function $y = 2e^x$: The derivative represents the slope of the tangent line at a given point.
   $\frac{dy}{dx} = \frac{d}{dx}[2e^x] = 2e^x$

2. Evaluate the slope at $x = 0$: Substitute $x = 0$ into the derivative to find the slope of the tangent line. $\text{slope} = 2e^0 = 2$

3. Write the equation of the tangent line: The equation of a tangent line is given by: $y - y_1 = m(x - x_1)$ where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency.
   Here, $m = 2$, $(x_1, y_1) = (0, 0)$. Substitute these values: $y - 0 = 2(x - 0)$ $y = 2x$

Final Answer:

The equation of the tangent line is: $y = 2x$

Would you like a detailed breakdown of any step, or have any further questions? Here are 5 related questions to explore:

1. What is the geometric significance of the tangent line to a curve at a point?
2. How does the derivative help in finding the slope of a tangent?
3. What happens if the function involves higher-order terms or products (e.g., $y = x^2e^x$)?
4. Can we find the normal line's equation at the same point? If so, how?
5. What is the slope of a tangent line to $y = 2e^x$ at $x = 1$?

Tip: For exponential functions like $e^x$, the slope of the tangent grows exponentially with $x$, reflecting the rapid growth of the function!