How do you model functions to fit given conditions?

  • Thread starter pakmingki
  • Start date
  • #1
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for example, i want to find the coeffiecients of a function f(x) = ax^2 + bx + c for these given conditions:
f(-1) = 3
f(1) = 3

i tried plugging and chugging
for the first condition:
3 = a - b + c
3 = a + b + c

now i tried subtracting the system to get
0 = 2b
so its solved that b = 0

so now we got a step further towards the goal, and we know we can right the function as
f(x) = ax^2 + c
so now we try plugging and chugging again, but then we get
3 = a + c
3 = a + c
since they are the same equation, i dont know how we can solve for a and c.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
lol you got the equation down to ax^2+c yes? Well for -1 and 1, x^2 is 1!

So now you have it down to a+c=3! No need to solve for, choose any values :D
1 and 2, 3 and 0, -4 and 7, whatever you want :D
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
962
for example, i want to find the coeffiecients of a function f(x) = ax^2 + bx + c for these given conditions:
f(-1) = 3
f(1) = 3
You have an obvious problem here- your quadratic, f(x)= ax2+ bx+ c, has three coefficients and you only have two conditions. Given three points, not on a straight line, there exist a unique quadratic passing through them- but there are an infinite number of parabolas passing through two given points.

i tried plugging and chugging
for the first condition:
3 = a - b + c
3 = a + b + c

now i tried subtracting the system to get
0 = 2b
so its solved that b = 0

so now we got a step further towards the goal, and we know we can right the function as
f(x) = ax^2 + c
so now we try plugging and chugging again, but then we get
3 = a + c
3 = a + c
since they are the same equation, i dont know how we can solve for a and c.
Yes, as I said, you don't have enough conditions. You can solve for c in terms of a: c= 3- a. All of the parabolas given by f(x)= ax2+ 3- a will pass through (1, 3) and (-1, 3).
 

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