# Suppose that the graph of f(x) is a parabola intersecting the origin and containing t

• Oct 4th 2009, 06:00 PM
Evan.Kimia
Suppose that the graph of f(x) is a parabola intersecting the origin and containing t
Suppose that the graph of f(x) is a parabola intersecting the origin and containing the points (-2,8) and (-3,6). Find f(x).

I think i start by setting the equation. A parabola is a quadratic eq. so id use ax^2+bx+c. Noticing that it intersects thru the origin, 0 is the y-in so its just ax^2+bx. Next I plug in the points -2,8 into the equation and do the same for -3,6 to get 2 diff. equations:

8=a(-2)^2+b(-2)
8=4a-2b
8-2b=4a
a=(8-2b)/4
-------------------------------------------------------------------
6=a(-3)^2+b(-3)
6=9a-3b
6-9a=-3b
(6-9a)/-3=b

-------------------------------------------------------------------
Im not sure if im right so far, or how to proceed. Any ideas?
Thanks.
• Oct 4th 2009, 06:13 PM
Jameson
This looks correct. You now have two variables, a and b, and two equations. Use substitution or another method to solve the system. I'm sure you've done this many times before, it just looks different because it's not using x and y as variables. Same idea though.