Question 610321
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi, 
what do the graphs for y=1/12(x)^2 and x = -1/4(y)^2 look like   ||Assume the + was an = sign
the vertex form of a parabola opening up or down, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex.The standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p)
{{{y=(1/12)(x)^2 }}} opens Upward as a = 1/12 > 0   Green
the vertex form of a parabola opening right or left, {{{x=a(y-k)^2 +h}}} where(h,k) is the vertex.The standard form is {{{(y -k)^2 = 4p(x -h)}}}, where  the focus is (h +p,k )
{{{x = (-1/4)(y)^2}}} opens to the LEFT as a = -1/4 < 0
{{{drawing(300,300,   -6, 6, -6, 6,  grid(1),
graph( 300, 300, -6, 6, -6, 6,0,(1/12)x^2, sqrt(-4x), -sqrt(-4x)))}}}