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

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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 $$ x = -\frac{11}{2} $$.
The quadratic function $$ f(x) = (x+4)(x+7) $$ has a minimum value of $$-\frac{9}{4}$$ at $$ x = -\frac{11}{2} $$.

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

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