SOLUTION: A graph of a parabola has a line of symmetry at x=3 and contains the points (5,-3) and (-1,9). Determine an equation for the parabola.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A graph of a parabola has a line of symmetry at x=3 and contains the points (5,-3) and (-1,9). Determine an equation for the parabola.       Log On


   

Question 378095: A graph of a parabola has a line of symmetry at x=3 and contains the points (5,-3) and (-1,9). Determine an equation for the parabola.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The vertex form is y+=+a%28x+-+h%29%5E2%2Bk. Since the line of symmetry is x = 3, then the x-ccordinate of the vertex is h = 3. Thus the equation now is y+=+a%28x-3%29%5E2+%2Bk.
We have to find a and k. Use the coordinates of the given point.
From (5, -3): -3+=+a%285-3%29%5E2+%2B+k, or -3 = 4a + k, or k = -4a-3.
From (-1, 9): 9+=+a%28-1-3%29%5E2+%2B+k, or 9 = 16a + k, or k = 9 - 16a.
Hence -4a - 3 = 9 - 16a.
12a = 12, or a = 1.
Then k = -4*1-3 = -7. Therefore the equation of the parabola is
y+=+%28x+-+3%29%5E2+-+7.