Answer
a. minimum
b. minimum value is $-\frac{3}{2}$ at x=$\frac{1}{2}$
c. domain = ($-\infty$,$\infty$) ; range = [$-\frac{3}{2},\infty$)
Work Step by Step
$f(x)=6x^{2}-6x$ ; a=6 b=-6
a. Because a>0, the function has a minimum value,
b. $x=-\frac{-b}{2a}$ $\rightarrow$ $x=-\frac{-6}{12}$ $\rightarrow$ $x=\frac{1}{2}$
$f(\frac{1}{2})=6(\frac{1}{2})^{2}-6(\frac{1}{2})$ = $-\frac{3}{2}$
Minimum value is $-\frac{3}{2}$ at $x=\frac{1}{2}$
c. Like all quadratic functions, the domain = ($-\infty$,$\infty$).
Range includes all real numbers at or above $-\frac{3}{2}$.
Therefore, the range = [$-\frac{3}{2},\infty$)
