Determine whether the quadratic function shown below has a minimum or maximum, then determine the minimum or maximum value of the function. \[ f(x)=(x+4)(x+7) \]

1. Expand the function: \[ f(x) = (x+4)(x+7) = x^2 + 11x + 28 \] 2. Identify the form: The quadratic function is in the form \( ax^2+bx+c \) where \( a = 1 \), \( b = 11 \), and \( c = 28 \). 3. Determine if the function has a minimum or maximum: Since \( a = 1 > 0 \), the parabola opens upward, which means the function has a minimum value. 4. Find the vertex: The \( x \)-coordinate of the vertex is given by: \[ x = -\frac{b}{2a} = -\frac{11}{2} \] Substitute \( x = -\frac{11}{2} \) into \( f(x) \) to find the minimum value: \[ f\left(-\frac{11}{2}\right) = \left(-\frac{11}{2}\right)^2 + 11\left(-\frac{11}{2}\right) + 28 \] Compute each term: \[ \left(-\frac{11}{2}\right)^2 = \frac{121}{4} \] \[ 11\left(-\frac{11}{2}\right) = -\frac{121}{2} \] Express all terms with denominator 4: \[ f\left(-\frac{11}{2}\right) = \frac{121}{4} - \frac{242}{4} + \frac{112}{4} \] Combine the terms: \[ f\left(-\frac{11}{2}\right) = \frac{121 - 242 + 112}{4} = \frac{-9}{4} \] 5. Conclusion: The function has a minimum value of \(\displaystyle -\frac{9}{4}\) at \( x = -\frac{11}{2} \).