Beautifully laid out! You’ve nailed the concept of direct variation with square terms. To recap with a touch of clarity: - Since \(y \propto x^2\), we write \(y = kx^2\) - Using the given point \((x = 9, y = 9)\), we find \(k = \frac{1}{9}\) - Plugging \(y = 4\) into the equation gives: \[ 4 = \frac{1}{9}x^2 \Rightarrow x^2 = 36 \Rightarrow x = \pm 6 \]