Answer
Standard form: $f(x)=(x+1)^2$
Vertex: $=(-1,0)$
Axis of symmetry: $x=-1$.
The x-intercept: $(-1,0)$
Work Step by Step
$f(x)=x^2+2x+1~~$ ($a=1,~b=2,~c=1$)
$f(x)=x^2+2(1)x+1^2$
$f(x)=(x+1)^2$
$-\frac{b}{2a}=-\frac{2}{2(1)}=-1$
$f(-1)=(-1+1)^2=0$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-1,0)$
So, the axis of symmetry is $x=-\frac{b}{2a}=-1$.
The x-intercepts:
$f(x)=0$
$(x+1)^2=0$
$x+1=0$
$x=-1$
So, the x-intercept point is: $(-1,0)$
