Answer
a. minimum
b. minimum is 11 at x=2
c. domain = (-$\infty$, $\infty$) ; range = [-11, $\infty$)
Work Step by Step
$f(x)=2x^{2}-8x-3$ ; a=2, b=-8, c=-3
a. Because a > 0, the function has a minimum value.
b. The minimum value occurs at $x=-\frac{b}{2a}$ $\rightarrow$ $x=-\frac{-8}{4}$ $\rightarrow$ x=2
$f(2)=2(2)^{2}-8(2)-3$ = -11
So the minimum value is -11 at x=2
c. Like all quadratic equations, the domain = (-$\infty$, $\infty$)
Because the function's minimum value is -11, the range includes all real numbers at or above -11. So the range = [-11,$\infty$)
