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We are given the function \( f(t) = -9t^2 - 3t + 7 \), and the task is to find:

1. \( f'(t) \), the first derivative of the function.
2. \( f'(1) \), the value of the derivative at \( t = 1 \).
3. \( f'(2) \), the value of the derivative at \( t = 2 \).

### Step 1: Find the derivative \( f'(t) \)
To differentiate \( f(t) = -9t^2 - 3t + 7 \), we apply basic rules of differentiation:

- The derivative of \( -9t^2 \) is \( -18t \) (using the power rule \( \frac{d}{dt}(t^n) = n t^{n-1} \)).
- The derivative of \( -3t \) is \( -3 \).
- The derivative of the constant \( 7 \) is \( 0 \).

Thus, the derivative \( f'(t) \) is:
\[
f'(t) = -18t - 3
\]

### Step 2: Find \( f'(1) \)
Substitute \( t = 1 \) into the derivative \( f'(t) = -18t - 3 \):
\[
f'(1) = -18(1) - 3 = -18 - 3 = -21
\]

### Step 3: Find \( f'(2) \)
Substitute \( t = 2 \) into the derivative:
\[
f'(2) = -18(2) - 3 = -36 - 3 = -39
\]

### Summary:
- \( f'(t) = -18t - 3 \)
- \( f'(1) = -21 \)
- \( f'(2) = -39 \)

Would you like more details on any part of this process?

Here are 5 related questions for further exploration:
1. What is the geometric interpretation of \( f'(t) \)?
2. How do second derivatives relate to concavity?
3. What is the equation of the tangent line at \( t = 1 \)?
4. Can we interpret \( f(t) \) in a real-world context, like physics?
5. How would the derivative change if \( f(t) \) had an additional cubic term?

**Tip:** The derivative gives the slope of the tangent line to the curve at any given point.

Find the first derivative of the function f(t) = -9t^2 - 3t + 7 and evaluate f'(1) and f'(2).

Learn how to differentiate the quadratic function f(t) = -9t^2 - 3t + 7 using the power rule, and evaluate the derivative at t=1 and t=2. Suitable for high school calculus students.

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