Answer
a) The function has a maximum value of $2$, and it occurs at $x=3$.
b) The domain of the function is $\left( -\infty ,\infty \right)$ and the range is $\left( -\infty ,\left. 2 \right] \right.$.
Work Step by Step
(a)
Compare the function $f\left( x \right)=-2{{x}^{2}}+12x-16$ with the standard parabolic equation $f\left( x \right)=a{{x}^{2}}+bx+c$; it is observed that:
$a=-2,b=12,c=-16$
Because $a<0$ , the parabola opens downwards and thereby the function has a maximum value.
Then, the coordinates of the vertex of the parabola will be the coordinates of the maxima of the parabola. Thus, the maximum value occurs at,
$x=-\frac{b}{2a}=-\frac{12}{2\left( -2 \right)}=\frac{12}{4}=3$
And, the maximum value of the function will be given as $f\left( 3 \right)$:
$\begin{align}
& f\left( 3 \right)=-2{{\left( 3 \right)}^{2}}+12\left( 3 \right)-16 \\
& \text{ }=-18+36-16 \\
& \text{ }=2
\end{align}$
Hence, the maximum value of the parabola occurs at $\left( 3,2 \right)$.
(b)
Since the leading coefficient of the quadratic function is negative thus the parabola opens downwards and its minimum value is $-\infty $. Also, from part (a), its maximum value is $-2$. Thus, the range of the function is $\left( -\infty ,-2 \right]$.
Since the function is defined for all real values of x, its domain is given by $\left( -\infty ,\infty \right)$.
Therefore, the domain of the function is $\left( -\infty ,\infty \right)$ and the range is $\left( -\infty ,\left. 2 \right] \right.$.
