Question 698279
{{{y=ax^2+bx+c }}}

Question: {{{y=x^2+2x+1}}}

To find the vertex of a parabola, we will write the function in the form {{{y=a(x-h)^2+k }}} where ({{{h}}},{{{k}}}) are the coordinates of vertex


{{{y=(x+1)^2}}}...=> {{{h=-1}}}, {{{k=0}}}; so, vertex is at ({{{-1}}},{{{0}}})

since {{{a=1}}}, parabola opens upward and has a minimum at {{{x=-1}}}

an {{{axis}}} of {{{symmetry}}} is the line that runs down its '{{{center}}} or {{{vertex}}}' and this line divides the graph into two perfect halves; so in your case that line is vertical line that goes through {{{x=-1}}}


range: all non-negative real numbers {{{y>=0}}}


{{{drawing(600,600,   -5, 5, -5, 10,  blue(line(-1,10,-1,-6)), grid(0),
graph( 600, 600, -5, 5, -5, 10,x^2+2x+1)) }}}