Find an equation of the parabola?
Find an equation of the parabola in standard for with vertex (-3,3), axis of symmetry y=3, and passing through the point (3,-1)
- 1 decade agoFavorite Answer
I don't think the first answer is correct. Since the axis of symmetry is y = 3, the parabola is of the form x = Ay^2 + By + C (that is, the parabola is C-shaped, not U-shaped). Since (3,-1) lies on the parabola, and the axis of symmetry is y = 3, the point (3, 7) also lies on the parabola. In addition the vertex (-3, 3) lies on the parabola. Substituting these three points into the equation above yields the three equations A - B + C = 3 and 49A + 7B + C = 3 and 9A + 3B + C = -3. Solve this system of equations to find A, B, and C and plug them into x = Ay^2 + By + C. The equation of the parabola will turn out to be x = (3/8)y^2 -(9/4)y + 3/8. The formula x = -b/2a, based on the concept of derivative (calculus), is invalid here since we are dealing with a C-shaped parabola.
- PuggyLv 71 decade ago
Note that the equation of a parabola in standard form is as follows.
y = A(x - h)^2 + k where the vertex is located at (h,k). Since our vertex is (-3,3), then
y = A(x + 3)^2 + 3
It passes through the point (3,-1), so we can plug those in for y and x to get
-1 = A(3 + 3) + 3
-1 = 6A + 3
-4 = 6A,
A = -4/6 = -2/3
Therefore, the parabola is
y = [-4/6](x + 3)^2 + 3
- Anonymous1 decade ago
The vertex is at x = -b/2a so -b/2a = -3. Then you have
-1 = 3*3*a + 3*b + c
3 = -3*-3*a + -3*b + c
-3 = -b/2a
so hopefully that's enough.