Answer
(a.) Minimum.
(b.) At $x=2$, minimum is $-13$.
(c.) Domain $=(-\infty,\infty)$.
Range $=[-13,\infty)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=3x^2-12x-1$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
$a=3,b=-12$ and $c=-1$
(a.)
Because $a>0$, the function has a minimum value.
(b.)
$x-$coordinate at which minimum value occurs is
$x=-\frac{b}{2a}$.
Substitute all values.
$x=-\frac{(-12)}{2(3)}$.
Simplify.
$x=2$.
Substitute the value of $x$ into the given function.
$f(2)=3(2)^2-12(2)-1$
Simplify.
$f(2)=12-24-1$
$f(2)=-13$
(c.)
Domain is all possible input values.
Domain $=(-\infty,\infty)$.
Range is all possible output values.
Minimum value is $-13$.
Range $=[-13,\infty)$.
