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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.