Expert: Sherry Wallin - 9/6/2009

Question
How would i write the equations of the parabolas with the following conditions:
a.  with vertex(5,-3) and passing through (3,9)

b.  passing through the points (-1,-6),(3,-12) and (2,-4.5)

Answer
Sorry Miranda for a late response. I wrote a solution and it went to website heaven...Here I will begin the problems and get you started...

For (a) you know that -b/2a = 5 -> -b = 10a and f(-b/2a) = -3 = a(-b/2a)^2 +b(-b/2) + c
You also know that f(3) = a(3^2) +b(3) + c = 9. And finally since (5,-3) is the vertex and parabolas are symmetric, the point (7,9) is also on the graph thus f(7) = a(7^2) + b(7) + c = 9. You have plenty of ammo to solve this system of equations.

For part b:
f(-1) = a(-1)^2 +b(-1) + c = -6 (eqn 1)
f(3) = a(3^2) +b(3) + c = -12  (eqn 2)
f(2) = a(2^2) +b(2) + c = -4.5  (eqn 3)

Again a system of equations with 3 equations and 3 unknowns.
a-b+ c = -6
-9a-3b-c = 12 (subtracted eqn 2 from eqn 1)
-8a-4b = 6 -> -4a-2b = 3 (divided through by 2) call this (eqn 4)
a-b+c = -6
-4a -2b -c = 4.5 (subtracted eqn 3 from eqn 1)
-3a -3b = -1.5 -> 2a + 2b = 1 (divided through by -1.5) call this eqn 5

-4a-2b = 3 eqn 4
2a + 2b = 1  eqn 5 add
-2a = 4 -> a = -2
2(-2) + 2b = 1 -> 2b = 5 -> b = 5/2 substitute into eqn 5 and solve for b
-2 -5/2 + c = -6 -> -9/2 + c = -12/2 -> c = -3/2 substitute a = -2 and b = 5/2 into eqn 1 to get c

y = -2x^2 + (5/2)x -3/2 is the equation for your parabola

Math Prof