Answer
(a.) Minimum.
(b.) At $x=2$, minimum is $-11$.
(c.) Domain $=(-\infty,\infty)$.
Range $=[-11,\infty)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=2x^2-8x-3$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
$a=2,b=-8$ and $c=-3$
(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.
$x=-\frac{(-8)}{2(2)}$.
Simplify.
$x=2$.
Substitute the value of $x$ into the given function.
$f(2)=2(2)^2-8(2)-3$
Simplify.
$f(2)=8-16-3$
$f(2)=-11$
(c.)
Domain is all possible input values.
Domain $=(-\infty,\infty)$.
Range is all possible output values.
Minimum value is $-11$.
Range $=[-11,\infty)$.
