Determine whether there is a maximum or minimum value for the given function, and find that value.
-$$f ( x ) = \frac { 1 } { 2 } x ^ { 2 } - 2 x - \frac { 15 } { 2 }$$
A) Minimum: $- \frac { 19 } { 2 }$
B) Maximum: 0
C) Minimum: $- \frac { 3 } { 2 }$
D) Maximum: - 9

Question

Quizplus · Accepted Answer

The given function is a quadratic function in the form $f(x) = ax^2 + bx + c$, where $a = \frac{1}{2}$, $b = -2$, and $c = -\frac{15}{2}$. Since $a > 0$, the parabola opens upwards, indicating a minimum value. To find the minimum value, we use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Substituting $b = -2$ and $a = \frac{1}{2}$, we get $x = -\frac{-2}{2(\frac{1}{2})} = 2$. Substituting $x = 2$ into the original equation to find the y-coordinate (minimum value), we get $f(2) = \frac{1}{2}(2)^2 - 2(2) - \frac{15}{2} = 2 - 4 - \frac{15}{2} = -\frac{19}{2}$.

Determine Whether There Is a Maximum or Minimum Value for the Given