To solve this problem, we need to graph the function $f(x)$ and the secant lines that pass through the point $P(4, 2)$ and $Q(x, f(x))$ for $x$-values of 3, 6, and 8. Additionally, we will calculate the slope of each secant line.

Step 1: Graph the Function $f(x)$

First, we need to know the function $f(x)$. Since it's not provided, I'll assume it's the function for which $P(4, 2)$ is a point on the graph. We can hypothesize that $f(x)$ could be a simple linear function $f(x) = x - 2$, which passes through $P(4, 2)$.

If $f(x) = x - 2$, then:

• $f(4) = 4 - 2 = 2$
• $f(3) = 3 - 2 = 1$
• $f(6) = 6 - 2 = 4$
• $f(8) = 8 - 2 = 6$

Step 2: Find the Slopes of the Secant Lines

The slope of a secant line passing through two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is calculated using the formula: $\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$

Let's calculate the slope for each secant line:

1. Secant line through $Q(3, f(3))$:

   • $x_1 = 4$, $y_1 = 2$
   • $x_2 = 3$, $y_2 = f(3) = 1$ $\text{slope} = \frac{1 - 2}{3 - 4} = \frac{-1}{-1} = 1.000$

2. Secant line through $Q(6, f(6))$:

   • $x_1 = 4$, $y_1 = 2$
   • $x_2 = 6$, $y_2 = f(6) = 4$ $\text{slope} = \frac{4 - 2}{6 - 4} = \frac{2}{2} = 1.000$

3. Secant line through $Q(8, f(8))$:

   • $x_1 = 4$, $y_1 = 2$
   • $x_2 = 8$, $y_2 = f(8) = 6$ $\text{slope} = \frac{6 - 2}{8 - 4} = \frac{4}{4} = 1.000$

Summary of Slopes

• The slope of the secant line passing through $Q(3, f(3))$: 1.000
• The slope of the secant line passing through $Q(6, f(6))$: 1.000
• The slope of the secant line passing through $Q(8, f(8))$: 1.000

These slopes indicate that if $f(x) = x - 2$, the function is linear and all the secant lines have the same slope of 1.

If you need to graph this or use a different function, please provide the exact form of $f(x)$.

Would you like more details or have any questions?

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Here are 5 related questions to consider:

1. What happens to the slope of the secant line as the two points $P$ and $Q$ get closer together?
2. How would the slopes change if the function $f(x)$ was quadratic instead of linear?
3. Can you find the derivative of $f(x)$ to compare with the slopes of the secant lines?
4. How does the choice of points affect the average rate of change represented by the secant line?
5. How would you graph these secant lines on a coordinate plane?

Tip: The slope of the secant line can be seen as the average rate of change of the function between the two points.