Consider the function $$ \mathrm{f}(\mathrm{x})=-2 \mathrm{x}^{2}+12 \mathrm{x}-8 $$. a. Determine, without graphing, whether the function has a minimum value or a maxime b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. a. The function has a b. The minimum/maximum value is $$ \square $$. It occurs at $$ \mathrm{x}=\square $$. c. The domain of f is $$ \square $$. (Type your answer in interval notation.) The range of f is $$ \square $$. (Type your answer in interval notation.)

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a. Since the coefficient of $$ x^2 $$ is $$-2$$, which is negative, the parabola opens downward. Therefore, the function has a maximum value.

b. To find the maximum value, we first determine the $$ x $$-coordinate of the vertex using  
$$
x = -\frac{b}{2a} = -\frac{12}{2(-2)} = 3.
$$
Substitute $$ x = 3 $$ into the function to get the maximum value:  
$$
f(3) = -2(3)^2 + 12(3) - 8 = -18 + 36 - 8 = 10.
$$
Thus, the maximum value is $$ 10 $$ and it occurs at $$ x = 3 $$.

c. The domain of $$ f $$ is all real numbers:  
$$
(-\infty, \infty).
$$
Since the parabola opens downward and the maximum value is $$ 10 $$ at $$ x=3 $$, the range is:  
$$
(-\infty, 10].
$$
a. The function has a maximum value. b. The maximum value is $$ 10 $$. It occurs at $$ x = 3 $$. c. The domain of $$ f $$ is $$ (-\infty, \infty) $$. The range of $$ f $$ is $$ (-\infty, 10] $$.

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