To find \( h \) for the function \(y = 1 + hx - 3x^2\) with a minimum value of 13: 1. The vertex \( x \) is given by \(x = \frac{h}{6}\) 2. Substitute into the function: \(y = 1 + \frac{h^2}{6} - 3\left(\frac{h}{6}\right)^2 = 1 + \frac{h^2}{6} - \frac{h^2}{12}\) 3. Combine terms: \(y = 1 + \frac{2h^2}{12} - \frac{h^2}{12} = 1 + \frac{h^2}{12}\) 4. Set equal to 13: \(1 + \frac{h^2}{12} = 13 \implies \frac{h^2}{12} = 12 \implies h^2 = 144 \implies h = 12 \text{ or } -12\) \( h = 12 \) or \( h = -12 \).