Answer
a. minimum
b. minimum value is $-\frac{5}{4}$ at $x=\frac{1}{2}$
c. domain = ($-\infty$,$\infty$). ; range = [$-\frac{5}{4},\infty$)
Work Step by Step
$f(x)=5x^{2}-5x$ ; a=5 b=-5
a. Because a>0, the function has a minimum value,
b. $x=-\frac{-b}{2a}$ $\rightarrow$ $x=-\frac{-5}{10}$ $\rightarrow$ $x=\frac{1}{2}$
$f(\frac{1}{2})=5(\frac{1}{2})^{2}-5(\frac{1}{2})$ = $-\frac{5}{4}$
Minimum value is $-\frac{5}{4}$ at $x=\frac{1}{2}$
c. Like all quadratic functions, the domain = ($-\infty$,$\infty$).
Range includes all real numbers at or above $-\frac{5}{4}$.
Therefore, the range = [$-\frac{5}{4},\infty$)
