Given that the quantity $y$ varies directly with the square of $x$, the relationship can be expressed as:

$y = kx^2$

where $k$ is the constant of proportionality.

Step 1: Find the constant $k$

We are given that $y = 16$ when $x = 2$. Substituting these values into the equation:

$16 = k(2^2)$

$16 = k(4)$

$k = \frac{16}{4} = 4$

Step 2: Use the constant $k$ to find $y$ when $x = 6$

Now, substitute $k = 4$ and $x = 6$ into the equation $y = kx^2$:

$y = 4(6^2)$

$y = 4(36) = 144$

Thus, $y = 144$ when $x = 6$.

Would you like further details or explanations on any step?

Here are 5 related questions for practice:

1. What happens to $y$ if $x$ triples in this type of relationship?
2. How would the equation change if $y$ varied inversely with $x^2$?
3. If $y$ is 64 when $x = 4$, what would $y$ be when $x = 8$?
4. How would you solve for $x$ if $y = 100$ and $k$ is known?
5. How do you derive the general equation for direct variation from experimental data?

Tip: In direct variation, as the independent variable increases, the dependent variable increases proportionally, but inversely if the variation is inverse.