Answer
The function has a minimum value.
The minimum value is $7$.
Work Step by Step
Comparing $f(x)=x^{2}+4x+11$ with $ax^{2}+bx+c$, we see that $a=1$ and $b=4$.
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
$x=-\frac{b}{2a}=-\frac{4}{2(1)}=-2$
Evaluating the function at $x=-2$, we get the y-coordinate of the vertex as
$f(-2)=(-2)^{2}+4(-2)+11=7$
