Use the quadratic function to determine if the function has a maximum or minimum value and then find this maximum or minimum value.
-$$f ( x ) = \frac { 1 } { 2 } x ^ { 2 } + 4 x - 4$$
A) minimum at $( - 4 , - 12 )$
B) maximum at $( 4,6 )$
C) minimum at $( 4,12 )$
D) maximum at $( - 4 , - 4 )$

Question

Quizplus · Accepted Answer

The quadratic function is in the form $f(x) = ax^2 + bx + c$, where $a = \frac{1}{2}$, $b = 4$, and $c = -4$. Since $a > 0$, the parabola opens upwards, indicating a minimum value. The vertex of the parabola, which gives the minimum value, can be found using the formula $-\frac{b}{2a}$ for the x-coordinate. Substituting $b = 4$ and $a = \frac{1}{2}$, we get $-\frac{4}{2(\frac{1}{2})} = -4$. To find the y-coordinate of the vertex, substitute $x = -4$ into the original equation: $f(-4) = \frac{1}{2}(-4)^2 + 4(-4) - 4 = 8 - 16 - 4 = -12$. Therefore, the function has a minimum at $(-4, -12)$.

Use the Quadratic Function to Determine If the Function Has f(x)=12x2+4x−4f ( x ) = \frac { 1 } { 2 } x ^ { 2 } + 4 x - 4f(x)=21​x2+4x−4

Use the Quadratic Function to Determine If the Function Has $f ( x ) = \frac { 1 } { 2 } x ^ { 2 } + 4 x - 4$