Find the equation of the parabola that contains these points?

The equation:

(x-f_x)^2+(y-f_y)^2 = (f_x (x-3 v_x)+f_y (y-3 v_y)+f_x^2+f_y^2+2 v_x^2-x v_x+2 v_y^2-y v_y)^2/(-2 f_x v_x-2 f_y v_y+f_x^2+f_y^2+v_x^2+v_y^2)

I used Wolfram|Alpha's computational math engine for this. Check it out below.

i think that the answer is (x^2-4) + 6
there are set groups of family graphs that corresponds to the types of graph like x^2 for parabolas ans x for lines, so in the equation x^2 say its a parabola and the -4 say that the graph is translated 4 to the right (+4 it moves to the left) then the graph is translated 6 units up because of +6 (-6 then 6 units down)
hope it helps:)

A quadratic is of the form ax^2 + bx + c= y

(1,9) ------> a(1)^2 + b(1) + c = 9

So a + b + c = 9

(4,6) ------> a(4)^2 + b(4) + c = 6

So 16a + 4b + c = 6

(6,14) -----> a(6)^2 + b(6) + c = 14

So 36a + 6b + c = 14


Now you have a system with three unknowns

a + b + c = 9---> (1)
16a + 4b + c = 6---> (2)
36a + 6b + c = 14--->(3)

Use elimination with 1 and 2.  Lets eliminate c, so multiply eqn 2 by -1

a + b + c = 9
-16a -4b -c = -6
_____________
-15a -3b = 3 ------>(4)

Use elimination with 1 and 3.  Lets again eliminate c, so multiply eqn 3 by -1


a + b + c = 9
-36a -6b - c = -14
________________
-35a - 5b = -5------>(5)


Now use your two new equations, 4 and 5, and solve.  Use substitution.

-15a -3b = 3
-35a - 5b = -5--------> b = -7a +1


-15a -3(-7a +1) = 3
-15a + 21a -3 = 3
6a -3 = 3
6a = 6
a = 1


b = -7(1) +1
b = -6

Now plug a and b into equation 1 to find c

1 + -6 + c = 9
-5 + c = 9
c = 14

So a = 1, b = -6, and c = 14

Plug those values into ax^2 + bx + c

x^2 -6x + 14

Not hard, but it takes a long time!

Hope this helps!

There are an infinite number of solutions.  The parabola could be aligned with the x-axis, the y-axis or tilted.

If the parabola is vertical it is of the form:
ax² + bx + c = y

Plug in the values of the points and get three equations in three unknowns.

a + b + c = 9
16a + 4b + c = 6
36a + 6b + c = 14

Solving we get:
a = 1
b = -6
c = 14

The equation of the parabola is:
y = x² - 6x + 14
__________

If the parabola is horizontal it is of the form:
ay² + by + c = x

Plug in the values of the points and get three equations in three unknowns.

81a + 9b + c = 1
36a + 6b + c = 4
196a + 14b + c = 6

Solving we get:
a = 1/4
b = -19/4
c = 47/2

The equation of the parabola is:
x = (1/4)y² - (19/4)y + 47/2

There are also an infinite number of tilted parabolas.