Question 92007
{{{f(x)= x^2+4x+1}}}

Let's find the vertex first

in standard format of quadratic equation {{{y = ax^2+bx+c}}}

the x value of vertex is given by {{{-b/(2a)}}}

in your function, a=1, b =4, so 

x = {{{-4/(2*1)}}} = -2

then f(-2) = -3

so vertex is (-2, -3)

Second part:

 X  ---------Y
        
-4  -------- 1
-3 -------- -2           
-2 -------- -3
-1 -------- -2
0  --------  1

from these numbers, you can see that they are symmetric, that's the characteristic of a parabola