The problem states:

We are given a point $P(2, 4)$ on the curve $y = x^2$, and another point $Q$ that also lies on this curve, with $Q \neq P$. The task is to express the slope of the line $PQ$ as a function of the $x$-coordinate of point $Q$, and sketch the graph of this slope function.

Step 1: Coordinates of Points

• Point $P$ has coordinates $(2, 4)$, as given.
• For point $Q$, its coordinates can be written as $(x_Q, y_Q)$, where $y_Q = x_Q^2$, since $Q$ also lies on the curve $y = x^2$.

Step 2: Formula for the Slope of Line $PQ$

The slope $m$ of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

In this case:

• $(x_1, y_1) = (2, 4)$ (point $P$)
• $(x_2, y_2) = (x_Q, x_Q^2)$ (point $Q$)

Thus, the slope of line $PQ$ is:

$m(x_Q) = \frac{x_Q^2 - 4}{x_Q - 2}$

Step 3: Simplifying the Expression

To simplify the slope function, we can factor the numerator:

$m(x_Q) = \frac{(x_Q - 2)(x_Q + 2)}{x_Q - 2}$

For $x_Q \neq 2$, the terms $(x_Q - 2)$ cancel out, leaving:

$m(x_Q) = x_Q + 2 \quad \text{for} \quad x_Q \neq 2$

Thus, the slope of the line $PQ$ is $m(x_Q) = x_Q + 2$.

Step 4: Conclusion and Graph

• The function for the slope of the line $PQ$ is $m(x_Q) = x_Q + 2$, which is a linear function.
• At $x_Q = 2$, the slope is undefined, as the two points coincide.

To sketch the graph of this function:

• The function is a straight line with a slope of 1 and a y-intercept at $y = 2$, excluding the point $x_Q = 2$ where the function is undefined.

Would you like more detailed steps or any additional clarifications?

Follow-up Questions:

1. How do we handle the slope at $x_Q = 2$ mathematically?
2. Can you explain the process of factoring the numerator more clearly?
3. What is the geometric interpretation of the line $PQ$ as $x_Q$ approaches 2?
4. How would this problem change if the curve equation was different?
5. What are the potential applications of finding the slope of secant lines?

Tip:

For problems involving secant lines, it helps to recall that as the two points $P$ and $Q$ get closer together, the slope of the secant line approaches the derivative of the function at that point.