Question 408858
  <pre><font size = 3 color = "indigo"><b>
Hi, Previously posted.
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
a. f (x) = x^2    |Parabola: Vertex Pt(0,0). y-axis the line of symmetry
{{{drawing(300,300,   -6, 6, -6, 6,  grid(1),
graph( 300, 300, -6, 6, -6, 6,x^2))}}}
b. f(x) = x^2+4x = (x+2)^2-4   |Parabola: Vertex Pt(-2,-4). x = -2 the line of symmetry
{{{drawing(300,300,   -6, 6, -6, 6,blue(line(-2,6,-2,-6)),     grid(1),
graph( 300, 300, -6, 6, -6, 6,x^2 + 4x))}}}
c. f(x) = x^2-4x +3 = (x-2)^2 -1 |Parabola: Vertex Pt(2,-1). x = 2 the line of symmetry
{{{drawing(300,300,   -6, 6, -6, 6,blue(line(2,6,2,-6)),   grid(1),
graph( 300, 300, -6, 6, -6, 6,x^2-4x+3))}}}