Part (a) - Secant Line Slope:

We are given the function $f(x) = x^2$, and we need to find the general equation for the slope of the secant line between the point with x-coordinate $a = 3$ and the point $(a + h, f(a + h))$.

The slope of a secant line between two points on a graph is given by:

$m_{\text{secant}} = \frac{f(a+h) - f(a)}{(a+h) - a}$

Here, $f(x) = x^2$, and the points are $(a, f(a)) = (3, f(3))$ and $(a+h, f(a+h)) = (3+h, f(3+h))$.

1. Step 1: Calculate $f(a) = f(3)$. $f(3) = 3^2 = 9$

2. Step 2: Calculate $f(a+h) = f(3+h)$. $f(3+h) = (3+h)^2 = 9 + 6h + h^2$

3. Step 3: Apply the secant line formula. $m_{\text{secant}} = \frac{f(3+h) - f(3)}{(3+h) - 3}$ Substituting the values: $m_{\text{secant}} = \frac{(9 + 6h + h^2) - 9}{h}$

4. Step 4: Simplify the expression. $m_{\text{secant}} = \frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h}$

5. Step 5: Factor out $h$ and simplify further. $m_{\text{secant}} = \frac{h(6 + h)}{h} = 6 + h$

Thus, the general equation for the slope of the secant line is: $m_{\text{secant}} = 6 + h$

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Part (b) - Estimate the Slope of the Tangent Line:

To estimate the slope of the tangent line at the point where $x = 3$, we can take the limit of the secant slope as $h \to 0$. The slope of the tangent line is:

$m_{\text{tangent}} = \lim_{h \to 0} (6 + h)$

Taking the limit: $m_{\text{tangent}} = 6 + 0 = 6$

Thus, the slope of the tangent line at $x = 3$ is: $m_{\text{tangent}} = 6$

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Do you have any questions or want more details on any of the steps?

Here are five related questions for further practice:

1. How do you find the slope of a secant line for a cubic function?
2. How does the difference between secant and tangent line slopes evolve as $h$ decreases?
3. What is the derivative of $f(x) = x^2$ at any point $x = a$?
4. What is the geometric interpretation of the tangent line on a curve?
5. How do limits apply in the definition of the derivative?

Tip: The slope of the tangent line at any point on a function represents the instantaneous rate of change at that point.