Here is the plot of the time response:

1. The top plot shows the input signal $x(t) = e^{2t} u(-t-2)$, which is only active for $t \leq -2$.
2. The middle plot shows the impulse response $h(t) = u(t-5)$, which starts at $t = 5$.
3. The bottom plot is the output response $y(t)$, obtained by convolving $x(t)$ and $h(t)$.

The output shows how the system responds to the input after processing through the convolution operation. Since the input is active for negative times and the impulse response starts at $t = 5$, the output begins after this delay.

Do you want further details or have any questions?

Here are five related questions:

1. How does the step function $u(t)$ affect the behavior of $h(t)$ and $x(t)$?
2. What happens if the impulse response is shifted differently?
3. How does convolution differ from multiplication of two signals?
4. Can you compute the convolution if the system is not time-invariant?
5. What is the effect of scaling $x(t)$ by a constant on the convolution result?

Tip: When performing convolution, always consider the time intervals where each function is active to simplify calculations.