Answer
The function has a minimum value.
The minimum value is $\frac{1}{2}$.
Work Step by Step
Comparing $y=2x^{2}-10x+13$ with $ax^{2}+bx+c$, we see that $a=2$ and $b=-10$.
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
$x=-\frac{b}{2a}=-\frac{-10}{2(2)}=\frac{5}{2}$
Evaluating the function at $x=\frac{5}{2}$, we get the y-coordinate of the vertex as
$y=2(\frac{5}{2})^{2}-10(\frac{5}{2})+13=\frac{1}{2}$
