Question 475403
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
Hi,
What point is the intersection of the graphs: 
x^2 + y = 2   
 y = -x^2 +2   Parabola V(0,2) opening downward (See below)
-y^2 + 3x = 2
 x = (1/3)y^2 + 2/3  Parabola V(2/3,0) opening to the right (See below)
y = -x        Line:  Pt(0,0) slope m = -1 (slants left)
algebraically:  substituting -x for y
x^2 -x - 2 = 0 = (x+1)(x-2)= 0,  x = - 1(Extraneous) and x = 2
 x = 2  then y = -2
Graphs: parabolas  and Line intersect at:(2,-2) 
{{{drawing(300,300,  -10,10,-10,10,     grid(1),circle(2/3, 0,0.3),
circle(2, -2,0.3),
graph( 300, 300,-10,10,-10,10,  0,-x,-x^2+2,sqrt(3x-2),-sqrt(3x-2) ))}}}

Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius

 Standard Form of an Equation of an Ellipse is {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1 }}}
where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

Standard Form of an Equation of an Hyperbola opening right and  left is:
  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} where Pt(h,k) is a center  with vertices 'a' units right and left of center.
Standard Form of an Equation of an Hyperbola opening up and down is:
  {{{(y-k)^2/b^2 - (x-h)^2/a^2 = 1}}} where Pt(h,k) is a center  with vertices 'b' units up and down from center.

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)

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 )