Answer
a) Maximum
b) The maximum value of the function is $1$ and it occurs at $x=1$.
c) Domain is $\left( -\infty ,\ \infty \right)$ and range is $\left( -\infty ,\ 1 \right]$
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: $f\left( x \right)=-4{{x}^{2}}+8x-3$. 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 the 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 the 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 $f\left( x \right)=-4{{x}^{2}}+8x-3$. Here $a<0$ and thus the function will have a maximum value.
Here, $a=-4,\ b=8$.
The function has a maximum at
$\begin{align}
& x=-\frac{b}{2a} \\
& =\frac{8}{8} \\
& =1.
\end{align}$
And the maximum value can be determined by calculating $f\left( 1 \right)$:
$\begin{align}
& f\left( 1 \right)=-4{{\left( 1 \right)}^{2}}+8\left( 1 \right)-3 \\
& =1.
\end{align}$
(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 $1$. Thus, the range of the function is $\left( -\infty ,\ 1 \right]$.
