<Question>
A curve is parameterized by
\[(x,y) = (t^3 + 7, -3t^2 - 6t - 5).\]Find the point the curve passes through at $t = 2.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the point that the curve passes through when t is 2. The curve is given in parametric equations: x is a function of t, which is t cubed plus 7, and y is another function of t, which is -3t squared minus 6t minus 5. Alright, let me think about how to approach this.

Since the problem is asking for the point on the curve at a specific value of t, which is 2, I remember that for parametric equations, you substitute the given t value into both the x and y equations and then compute the resulting coordinates. So, basically, plug t = 2 into both x(t) and y(t), and that should give me the (x, y) coordinates of the point.

Let me start with the x-component. The x equation is x = t³ + 7. Substituting t = 2 here, that would be x = (2)³ + 7.
Calculating 2 cubed, which is 2 * 2 * 2, that's 8. Then adding 7 to that gives 8 + 7 = 15. So, the x-coordinate when t = 2 is 15.

Now, moving on to the y-component. The y equation is y = -3t² - 6t - 5. Again, substituting t = 2 into this equation. Let's break it down step by step to avoid mistakes. First, compute t squared: (2)² is 4. Then multiply that by -3: -3 * 4 = -12. Next, compute the linear term: -6t, which is -6 * 2 = -12. Then, there's the constant term, which is -5. So, adding all those together: -12 -12 -5.

<prompt_1.1> Follow the above plan to calculate the y-coordinate by substituting t = 2 into the y equation and solve step by step. </prompt_1.1>
-----

Is there any chance that I might have mixed up the order of the components? For example, sometimes in parametric equations, people might use different parameters or mix up x and y. Let me check the original problem again.

<prompt_1.2> Check the original problem to confirm the parametric equations and ensure no mix-up in the order of components. </prompt_1.2>

Another thing to verify: did I substitute t = 2 into both equations correctly? Let's re-express the equations with t = 2.

<prompt_1.3> Re-express the parametric equations with t = 2 to verify the substitution and confirm the correctness of the solution. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me add those: -12 minus 12 is -24, then minus 5 more is -29. Wait, that seems straightforward, but let me double-check each step to make sure I didn't make an arithmetic error. 

First term: -3*(2)^2. The exponent comes first, so 2 squared is 4. Multiply by -3: -3*4 is indeed -12. Second term: -6*2. That's -12. Third term is -5. So adding them: -12 -12 is -24, then -24 -5 is -29.
</execution_1.1>

<execution_1.2>
The problem states the curve is parameterized by (x, y) = (t³ + 7, -3t² - 6t - 5). So, the first component is x, the second is y. Therefore, the answer should be (15, -29).
</execution_1.2>

<execution_1.3>
For x: (2)^3 + 7 = 8 + 7 = 15. For y: -3*(2)^2 -6*(2) -5. Let's compute each term again:

First term: -3*(4) = -12. Second term: -6*2 = -12. Third term: -5. Adding them: -12 -12 = -24; -24 -5 = -29.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, that seems correct, but let me check once more.
Alternatively, maybe I can compute the y-coordinate again step by step:

<prompt_2.1> Re-calculate the y-coordinate using the original substitution to ensure accuracy. </prompt_2.1>

Based on execution_1.1, execution_1.3:
Therefore, the point on the curve when t = 2 is (15, -29).
I think that's it. There's not much more to it since it's a straightforward parametric curve evaluation. But just to be thorough, maybe I can also check if I substituted the value correctly into the equations.

<prompt_2.2> Perform a final verification of the substitution and conclude the solution. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Compute t squared: 2*2=4, 4*2=8. So t²=4. Multiply by -3: 4*3=12, so -12. Then -6*2: 6*2=12, so -12. Then -5. So total is -12 -12 -5. Yes, that adds up to -29.
</execution_2.1>

<execution_2.2>
For x(t), it's t³ + 7. So plugging in 2: 2³ is 8, plus 7 is 15. For y(t), it's -3t² -6t -5. Plugging in 2: -3*(2)^2 is -12, -6*(2) is -12, minus 5. So total is -12 -12 -5 = -29. Yep, that's correct. 

I don't see any mistakes here. The calculations seem straightforward, but it's always good to verify each step. Maybe I can also visualize the curve a bit, but since we're only asked for the point at t=2, that's not necessary. However, just to satisfy my curiosity, maybe plotting a few points would help ensure that the parametrization makes sense.
</execution_2.2>
</Execution_2>

<Final_answer>
Therefore, the point the curve passes through at $t = 2$ is $\boxed{(15, -29)}$.
</Final_answer>
</think>