Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then
find that value.
-$$f ( x ) = 4 x ^ { 2 } + 12 x$$
A) minimum; $- \frac { 3 } { 2 }$
B) minimum; $- 9$
C) maximum; $- \frac { 3 } { 2 }$
D) maximum; $- 9$

Question

Quizplus · Accepted Answer

The quadratic function is in the form $f(x) = ax^2 + bx + c$, where $a = 4$, $b = 12$, and $c = 0$. Since $a > 0$, the parabola opens upwards, indicating a minimum value. To find the minimum value, use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Substituting $a = 4$ and $b = 12$ gives $x = -\frac{12}{2(4)} = -\frac{12}{8} = -\frac{3}{2}$. Then, substitute $x = -\frac{3}{2}$ into the original equation to find the minimum value: $f(-\frac{3}{2}) = 4(-\frac{3}{2})^2 + 12(-\frac{3}{2}) = 4(\frac{9}{4}) - 18 = 9 - 18 = -9$.

Determine, Without Graphing, Whether the Given Quadratic Function Has a Maximum