## Algebra 2 Common Core

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