Answer
a) Maximum
b) The maximum value of the function is $-3$ and it occurs at $x=21$.
c) Domain is $\left( -\infty ,\ \infty \right)$ and range is $\left( -\infty ,\ \ 21 \right]$
Work Step by Step
(a)
Let us consider a quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$.
If the constant $a>0,$ the parabola opens upward and the vertex is at its lowest point and the function would have a minimum value.
If the constant $a<0,$ the parabola opens downward and the vertex is at its highest point and the function would have a maximum value.
Also, consider the given function: $f\left( x \right)=-2{{x}^{2}}-12x+3$. Here $a<0$ and thus the function will have a maximum value.
(b)
let us consider a quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$.
If the constant $a>0,$ then function has a minimum that occurs at $x=-\frac{b}{2a}$ and the minimum value is $f\left( -\frac{b}{2a} \right)$.
If the constant $a<0,$ then function has a maximum that occurs at $x=-\frac{b}{2a}$ and the minimum value is $f\left( -\frac{b}{2a} \right)$.
Also, consider the given function: $f\left( x \right)=-2{{x}^{2}}-12x+3$.
Here $a<0$ and thus the function will have a maximum value.
Here, $a=-2,\ b=-12$.
The function has a maximum at:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\left( \frac{-12}{-4} \right) \\
& =-3
\end{align}$
And the maximum value can be determined by calculating $f\left( -3 \right):$
$\begin{align}
& f\left( -3 \right)=-2{{\left( -3 \right)}^{2}}-12\left( -3 \right)+3 \\
& =21.
\end{align}$
(c)
So, the function corresponds to every input on $x-$ axis as it widens at both ends. Thus, the domain of the function is $\left( -\infty ,\ \infty \right)$.
The vertex is the highest point of the function and all the outputs fall at or below $21$. Hence, the range of the function is $\left( -\infty ,\ 21 \right]$.
