Determine whether the given quadratic function has a minimum value or maximum value. Then find the coordinates of
the minimum or maximum point.
-$$f ( x ) = 4 x ^ { 2 } - 12 x$$
A) minimum; $\left( - \frac { 3 } { 2 } , - 9 \right)$
B) maximum; $\left( \frac { 3 } { 2 } , - 9 \right)$
C) minimum; $\left( \frac { 3 } { 2 } , - 9 \right)$
D) maximum; $\left( - \frac { 3 } { 2 } , - 9 \right)$

Question

Quizplus · Accepted Answer

The quadratic function $f(x) = 4x^2 - 12x$ opens upwards because the coefficient of $x^2$ is positive, indicating a minimum value. The vertex (minimum point) can be found using the formula $x = -\frac{b}{2a}$, giving $x = \frac{3}{2}$. Substituting $x = \frac{3}{2}$ into the function gives the minimum value $f\left(\frac{3}{2}\right) = -9$, so the correct answer is minimum; $\left(\frac{3}{2}, -9\right)$.

Determine Whether the Given Quadratic Function Has a Minimum Value f(x)=4x2−12xf ( x ) = 4 x ^ { 2 } - 12 xf(x)=4x2−12x

Determine Whether the Given Quadratic Function Has a Minimum Value $f ( x ) = 4 x ^ { 2 } - 12 x$