The problem provides two points, $A(-2, -4)$ and $B(4, 6)$, and asks for the value of $k$ such that the point $(k, 3)$ lies on the continuous curve AB.

Assuming that the curve is a straight line between points A and B (since no other information about the curve is provided), we can find the equation of the line passing through points A and B. After that, we can substitute $y(k) = 3$ to find the corresponding $k$-value.

Step 1: Find the slope of the line

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

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

For points $A(-2, -4)$ and $B(4, 6)$:

$m = \frac{6 - (-4)}{4 - (-2)} = \frac{6 + 4}{4 + 2} = \frac{10}{6} = \frac{5}{3}$

Step 2: Equation of the line

Using the point-slope form of the equation of a line, which is:

$y - y_1 = m(x - x_1)$

Substitute $m = \frac{5}{3}$ and point $A(-2, -4)$:

$y - (-4) = \frac{5}{3}(x - (-2))$ $y + 4 = \frac{5}{3}(x + 2)$

Simplify:

$y + 4 = \frac{5}{3}x + \frac{10}{3}$ $y = \frac{5}{3}x + \frac{10}{3} - 4$ $y = \frac{5}{3}x - \frac{12}{3}$ $y = \frac{5}{3}x - 4$

Step 3: Find $k$ such that $y(k) = 3$

Substitute $y = 3$ into the equation:

$3 = \frac{5}{3}k - 4$

Now, solve for $k$:

$3 + 4 = \frac{5}{3}k$ $7 = \frac{5}{3}k$ $k = \frac{7 \times 3}{5} = \frac{21}{5}$

Thus, $k = \frac{21}{5}$ or 4.2.

Final Answer:

The value of $k$ is $\frac{21}{5}$ or approximately 4.2.

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would the process change if the curve between A and B was not a straight line?
2. What is the importance of the slope in determining the equation of a line?
3. How do you find the distance between two points in a plane?
4. Can you express the equation of the line in standard form?
5. How would the answer change if $y(k)$ were a different value?

Tip: Always check your line equation by substituting the original points to confirm it's correct!