Answer
a. minimum
b. maximum value is 1 at x=1.
c. domain = (-$\infty$, $\infty$) ; range = ($-\infty$, 1]
Work Step by Step
$f(x)=-4x^{2}+8x-3$ ; a=-4, b=8, c=-3
a. Because a< 0, the function has a maximum value.
b. The maximum value occurs at $x=-\frac{b}{2a}$ $\rightarrow$ $x=-\frac{8}{-8}$ $\rightarrow$ x=1
$f(1)=-4(1)^{2}+8(1)-3$ = 1
So the maximum value is 1 at x=1
c. Like all quadratic equations, the domain = (-$\infty$, $\infty$)
Because the function's maximum value is 1, the range includes all real numbers at or below 1. So the range = ($-\infty$, 1]
