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Let's work through the problem shown in the image.

### Problem Summary:
We are given a function \( f(x) = x^2 - 1 \) and the point \( P(3, 8) \) on the graph of this function. The problem has three parts:

1. **Find the slope of the tangent line at \( P \) using the definition of the derivative:**

   \[
   m_{\text{tan}} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
   \]

   where \( f(a) \) is the value of the function at point \( a \).

2. **Determine the equation of the tangent line at \( P \).**

3. **Plot the graph of the function \( f(x) \) and the tangent line at \( P \).**

### Part (a) - Finding the slope of the tangent line at \( P(3, 8) \)

The formula to find the slope of the tangent line at a point is the definition of the derivative:
\[
m_{\text{tan}} = \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3}
\]

For \( f(x) = x^2 - 1 \), we first calculate \( f(3) \):
\[
f(3) = 3^2 - 1 = 9 - 1 = 8
\]

Now, substitute \( f(x) = x^2 - 1 \) and \( f(3) = 8 \) into the formula:
\[
m_{\text{tan}} = \lim_{x \to 3} \frac{(x^2 - 1) - 8}{x - 3}
\]
Simplify the numerator:
\[
m_{\text{tan}} = \lim_{x \to 3} \frac{x^2 - 9}{x - 3}
\]

Notice that \( x^2 - 9 \) is a difference of squares, so it factors as:
\[
m_{\text{tan}} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}
\]

Cancel out the \( (x - 3) \) terms:
\[
m_{\text{tan}} = \lim_{x \to 3} (x + 3)
\]

Now, substitute \( x = 3 \) into the remaining expression:
\[
m_{\text{tan}} = 3 + 3 = 6
\]

So, the slope of the tangent line at \( P \) is **6**.

### Part (b) - Determine the equation of the tangent line at \( P \)

The equation of a line is given by the point-slope form:
\[
y - y_1 = m(x - x_1)
\]

Here, the point \( P \) is \( (3, 8) \) and the slope \( m_{\text{tan}} = 6 \). Substituting these values into the point-slope form:
\[
y - 8 = 6(x - 3)
\]
Simplify the equation:
\[
y - 8 = 6x - 18
\]
\[
y = 6x - 10
\]

So, the equation of the tangent line is:
\[
y = 6x - 10
\]

### Part (c) - Plot the graph of \( f(x) = x^2 - 1 \) and the tangent line

To plot the function \( f(x) = x^2 - 1 \), we can sketch a parabola with its vertex at \( (0, -1) \) and its tangent line at \( P(3, 8) \), with the equation \( y = 6x - 10 \).

Would you like me to create a graph for you or answer any questions regarding the solution?

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Here are some related questions for deeper understanding:
1. What is the geometric meaning of the derivative at a point on a curve?
2. How would you find the derivative of a more complex function, like a cubic function?
3. Can the slope of a tangent line be negative? If so, when?
4. How does the definition of a derivative relate to the concept of limits?
5. What happens to the tangent line if we choose different points on the parabola?

**Tip:** When simplifying rational functions involving limits, look for factoring opportunities such as the difference of squares or cubes to help cancel terms effectively.

Given the function f(x) = x^2 - 1 and point P(3, 8), (a) use the definition of the derivative to find the slope of the tangent line at P, (b) determine the equation of the tangent line at P, and (c) plot the graph of f(x) and the tangent line.

Learn how to find the slope of the tangent line to f(x) = x^2 - 1 at P(3, 8) using the definition of the derivative. This problem involves understanding limits, derivatives, and quadratic functions. A detailed solution walks through finding the equation of the tangent line and graphing it.

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