Answer
a. minimum
b. minimum is -13 at x=2
c. domain = (-$\infty$, $\infty$) range = [-13, $\infty$)
Work Step by Step
$f(x)=3x^{2}-12x-1$ ; a=3, b=-12, c=-1
a. Because a > 0, the function has a minimum value.
b. The minimum value occurs at $x=-\frac{b}{2a}$ $\rightarrow$ $x=-\frac{-12}{6}$ $\rightarrow$ x=2
$f(2)=3(2)^{2}-12(2)-1$ = -13
So the minimum value is -13 at x=2
c. Like all quadratic equations, the domain = (-$\infty$, $\infty$)
Because the function's minimum value is -13, the range includes all real numbers at or above -13. So the range = [-13,$\infty$)
