Answer
a) Minimum
b) The minimum value of the function is $-\frac{3}{2}$ and it occurs at $x=\frac{1}{2}$.
c) Domain is $\left( -\infty ,\ \infty \right)$ and range is $\left[ -\frac{3}{2},\ \infty \right)$.
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 $f\left( x \right)=6{{x}^{2}}-6x$.
Here $a>0$ and thus 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 $f\left( -\frac{b}{2a} \right)$.
Also, consider the given function $f\left( x \right)=6{{x}^{2}}-6x$. Here $a>0$ and thus the function will have a minimum value.
Here, $a=6,\ b=-6.$.
The function has a minimum at
$\begin{align}
& x=-\frac{b}{2a} \\
& =\frac{6}{12} \\
& =\frac{1}{2}
\end{align}$
And the minimum value can be determined by calculating $f\left( \frac{1}{2} \right)$:
$\begin{align}
& f\left( \frac{1}{2} \right)=6{{\left( \frac{1}{2} \right)}^{2}}-6\left( \frac{1}{2} \right) \\
& =-\frac{3}{2}.
\end{align}$
(c)
And the function corresponds to every input on $x$ -axis as it widens at both ends. Thus, the domain of the function is $\left( -\infty ,\ \infty \right)$.
Thus, the vertex is the lowest point of the function and all the outputs fall at or above $-11$. Hence, the range of the function is $\left[ -\frac{3}{2},\ \infty \right)$.
