Answer
(a.) Minimum.
(b.) At $x=\frac{1}{2}$, minimum value is $-\frac{5}{4}$.
(c.) Domain $=(-\infty,\infty)$.
Range $=\left[-\frac{5}{4},\infty\right) $.
Work Step by Step
The given function is a quadratic function:
$f(x)=5x^2-5x$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
$a=5,b=-5$ and $c=0$
(a.)
Because $a>0$, the function has a minimum value.
(b.)
$x-$coordinate at which minimum value occurs is
$x=-\frac{b}{2a}$.
Substitute all values.
$x=-\frac{(-5)}{2(5)}$.
Simplify.
$x=\frac{1}{2}$.
Substitute the value of $x$ into the given function.
$f\left(\frac{1}{2}\right)=5\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)$
Simplify.
$f\left(\frac{1}{2}\right)=\frac{5}{4}-\frac{5}{2}$
The LCD is $4$.
Multiply the numerator and the denominator to form LCD at the denominator.
$f\left(\frac{1}{2}\right)=\frac{5}{4}-\frac{10}{4}$
Add numerators.
$f(\frac{1}{2})=\frac{5-10}{4}$
Simplify.
$f\left(\frac{1}{2}\right)=-\frac{5}{4}$
(c.)
Domain is all possible input values.
Domain $=(-\infty,\infty)$.
Range is all possible output values.
Minimum value is $-\frac{5}{4}$.
Range $=\left[-\frac{5}{4},\infty\right)$.
