Answer
a) Minimum
b) The minimum value of the function is $-13$ and it occurs at $x=2$.
c) Domain is $\left( -\infty ,\ \infty \right)$ and range is $\left[ -13,\ \infty \right)$
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: $f\left( x \right)=3{{x}^{2}}-12x-1$. Here, $a>0$ and Therefore, the function will have a minimum 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 $f\left( -\frac{b}{2a} \right)$.
And consider the given function $f\left( x \right)=3{{x}^{2}}-12x-1$. Here, $a>0$ and thus, the function will have a minimum value.
Here, $a=3,\ b=-12$.
The function has a minimum at
$\begin{align}
& x=-\frac{b}{2a} \\
& =\frac{12}{6} \\
& =2.
\end{align}$
And the minimum value can be determined by calculating $f\left( 2 \right)$:
$\begin{align}
& f\left( 2 \right)=3{{\left( 2 \right)}^{2}}-12\left( 2 \right)-1 \\
& =-13.
\end{align}$
(c)
The function corresponds to every input on the $x$ -axis as it widens at both ends. Thus, the domain of the function is $\left( -\infty ,\ \infty \right)$.
The vertex is the lowest point of the function and all the outputs fall at or above $-13$. Thus, the range of the function is $\left[ -13,\ \infty \right)$.
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 $f\left( -\frac{b}{2a} \right)$.
And consider the given function $f\left( x \right)=3{{x}^{2}}-12x-1$. Here, $a>0$ and thus, the function will have a minimum value.
Here, $a=3,\ b=-12$.
The function has a minimum at
$\begin{align}
& x=-\frac{b}{2a} \\
& =\frac{12}{6} \\
& =2.
\end{align}$
And the minimum value can be determined by calculating $f\left( 2 \right)$:
$\begin{align}
& f\left( 2 \right)=3{{\left( 2 \right)}^{2}}-12\left( 2 \right)-1 \\
& =-13.
\end{align}$
(c)
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 lowest point of the function and all the outputs fall at or above $-13$. Thus, the range of the function is $\left[ -13,\ \infty \right)$.
