#### Answer

If $y$ varies directly with $x^{2}$, and $y=2$ when $x=4$, then $y=\frac{81}{8}$ when $x=9$.

#### Work Step by Step

The direct variation function for this problem must be modified to $\frac{y}{x^{2}}=k$ (the $x$ variable is squared), since the problem states that y varies directly with $x^{2}$, not just $x$.
Using two forms of $\frac{y}{x^{2}}$, set up a proportion to find what $y$ equals when $x=9$.
$\frac{2}{4^{2}}=\frac{y}{9^{2}}$
Now write the cross products
$(4^{2})y=2(81)$
Simplify
$16y=162$
Divide both sides by 16
$\frac{16y}{16}=\frac{162}{16}$
Simplify
$y=\frac{81}{8}$