Answer
The function has a minimum value.
The minimum value is $-\frac{17}{4}$.
Work Step by Step
Comparing $f(x)=\frac{1}{5}x^{2}-5x+27$ with $ax^{2}+bx+c$, we see that $a=\frac{1}{5}$ and $b=-5$.
As $a\gt0$, the parabola opens up and the function has a minimum value. The minimum value is the y-coordinate of the vertex.
The x-coordinate of the vertex is given by
$x=-\frac{b}{2a}=-\frac{-5}{2(\frac{1}{5})}=\frac{25}{2}$
Evaluating the function at $x=\frac{25}{2}$, we get the y-coordinate of the vertex as
$f(\frac{25}{2})=\frac{1}{5}(\frac{25}{2})^{2}-5(\frac{25}{2})+27=-\frac{17}{4}$
