To solve the problem, we need to find the value of \( h \) such that the minimum value of the quadratic function \( y = 1 + hx - 3x^2 \) is equal to 13. Let's break this down step-by-step. ### Step 1: Understanding the Quadratic Function The given function is: \[ y = 1 + hx - 3x^2 \] This is a quadratic function in the standard form \( y = ax^2 + bx + c \), where: - \( a = -3 \) (the coefficient of \( x^2 \)), - \( b = h \) (the coefficient of \( x \)), - \( c = 1 \) (the constant term). ### Step 2: Finding the Vertex of the Quadratic For a quadratic function \( y = ax^2 + bx + c \), the vertex (which gives the minimum or maximum value) occurs at: \[ x = -\frac{b}{2a} \] In our case: - \( a = -3 \) - \( b = h \) Substituting these values into the vertex formula gives: \[ x = -\frac{h}{2(-3)} = \frac{h}{6} \] ### Step 3: Calculating the Minimum Value of \( y \) Now, we need to find the minimum value of \( y \) at \( x = \frac{h}{6} \). We substitute \( x = \frac{h}{6} \) back into the equation for \( y \): \[ y = 1 + h\left(\frac{h}{6}\right) - 3\left(\frac{h}{6}\right)^2 \] Calculating each term: 1. The first term is \( 1 \). 2. The second term is \( \frac{h^2}{6} \). 3. The third term is: \[ -3\left(\frac{h}{6}\right)^2 = -3 \cdot \frac{h^2}{36} = -\frac{h^2}{12} \] Now, combining these terms: \[ y = 1 + \frac{h^2}{6} - \frac{h^2}{12} \] ### Step 4: Simplifying the Expression To combine \( \frac{h^2}{6} \) and \( -\frac{h^2}{12} \), we need a common denominator. The least common multiple of 6 and 12 is 12: \[ \frac{h^2}{6} = \frac{2h^2}{12} \] Thus, we can rewrite \( y \): \[ y = 1 + \frac{2h^2}{12} - \frac{h^2}{12} = 1 + \frac{h^2}{12} \] ### Step 5: Setting the Minimum Value Equal to 13 According to the problem, the minimum value of \( y \) is 13: \[ 1 + \frac{h^2}{12} = 13 \] Subtracting 1 from both sides gives: \[ \frac{h^2}{12} = 12 \] ### Step 6: Solving for \( h^2 \) Multiplying both sides by 12: \[ h^2 = 144 \] Taking the square root of both sides: \[ h = 12 \quad \text{(since \( h \) is typically considered positive in this context)} \] ### Conclusion Thus, the value of \( h \) that satisfies the condition is: \[ \boxed{12} \] ### Explanation of Other Options - **Option A (13)**: This value does not satisfy the equation \( 1 + \frac{h^2}{12} = 13 \) because it would imply \( h^2 = 144 \) which is not true for \( h = 13 \). - **Option C (11)**: If \( h = 11 \), then \( \frac{h^2}{12} = \frac{121}{12} \), which does not equal 12 when added to 1. - **Option D (10)**: Similarly, if \( h = 10 \), then \( \frac{h^2}{12} = \frac{100}{12} \), which also does not satisfy the equation. ### Revision Summary - The minimum value of a quadratic function occurs at the vertex, calculated using \( x = -\frac{b}{2a} \). - Substitute the vertex \( x \) back into the function to find the minimum value. - Set the minimum value equal to the given value (13) and solve for \( h \). - The correct answer is \( h = 12 \).