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### Given Function:
\[ f(x) = -\frac{5}{2}x^2 + 8x + 8 \]

### 1. Find \( f'(x) \) (the derivative of \( f(x) \)):
We differentiate the function term by term:
\[
f'(x) = \frac{d}{dx}\left( -\frac{5}{2}x^2 + 8x + 8 \right)
\]
- The derivative of \( -\frac{5}{2}x^2 \) is \( -5x \) (using the power rule).
- The derivative of \( 8x \) is \( 8 \).
- The derivative of the constant \( 8 \) is \( 0 \).

Thus, 
\[
f'(x) = -5x + 8
\]

### 2. Find \( f'(-1) \):
Substitute \( x = -1 \) into \( f'(x) \):
\[
f'(-1) = -5(-1) + 8 = 5 + 8 = 13
\]

### 3. Find the slope and equation of the tangent line at \( (-1, -\frac{5}{2}) \):
- The slope of the tangent line is given by \( f'(-1) \), which we calculated to be **13**.

Next, we find the equation of the tangent line using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
Where:
- \( m = 13 \) (the slope),
- \( (x_1, y_1) = \left(-1, -\frac{5}{2}\right) \) (the given point on the graph).

Substitute the values:
\[
y - \left(-\frac{5}{2}\right) = 13(x - (-1))
\]
Simplify:
\[
y + \frac{5}{2} = 13(x + 1)
\]
\[
y + \frac{5}{2} = 13x + 13
\]
Now subtract \( \frac{5}{2} \) from both sides to get the equation in slope-intercept form:
\[
y = 13x + 13 - \frac{5}{2}
\]
Convert \( 13 \) to a fraction:
\[
y = 13x + \frac{26}{2} - \frac{5}{2}
\]
\[
y = 13x + \frac{21}{2}
\]
Thus, the equation of the tangent line is:
\[
y = 13x + \frac{21}{2}
\]

### Final Answers:
1. \( f'(x) = -5x + 8 \)
2. \( f'(-1) = 13 \)
3. Slope = 13, Equation of tangent line: \( y = 13x + \frac{21}{2} \)

Would you like any further details or explanations?

Here are 5 related questions for practice:
1. What is the second derivative \( f''(x) \) for the given function?
2. How do you find the critical points of \( f(x) \)?
3. What is the concavity of \( f(x) \) at \( x = -1 \)?
4. How does the equation of the normal line at the point \( (-1, -\frac{5}{2}) \) differ from the tangent line?
5. What is the vertex of the parabola represented by \( f(x) \)?

**Tip**: The tangent line at any point on a curve has the same slope as the derivative of the function at that point.

Consider the following: f(x) = -5/2x^2 + 8x + 8. Find f'(x), f'(-1), and the slope and equation of the tangent line to the graph of the function f at the point (-1, -5/2).

This problem involves finding the derivative of a quadratic function, calculating the slope of the tangent line at a specific point, and determining the equation of the tangent line using the point-slope form. This is a fundamental problem in introductory calculus.

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