#### Answer

$A$

#### Work Step by Step

RECALL:
(1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens:
(i) upward when $a \gt0$ and has its vertex as its minimum; or
(ii) downward when $a\lt0$ and has its vertex as its maximum.
(2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$
Let's compare $f(x)=x^2+2x+1$ to $f(x)=ax^2+bx+c$.
We can see that $a=1$, $b=2$, $c=1$.
$a\gt0$, hence the graph opens up, hence it's vertex is a minimum.
The minimum value is at $x=-\frac{b}{2a}=-\frac{2}{2\cdot(1)}=-1.$
Hence the minimum value is $f(-1)=(-1)^2+2(-1)+1=0.$
Thus, the vertex is at $(-1,0)$ and the graph opens up.
Therefore, the graph of the given function is the one in graph $A$.