SOLUTION: Given a table of x and F(x) values how would you find the quadratic equation that describes it? x -4, -3, -2, -1, 0 y -1, 0, 3, 8, 15 Since when x=0, y=15; I know t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Given a table of x and F(x) values how would you find the quadratic equation that describes it? x -4, -3, -2, -1, 0 y -1, 0, 3, 8, 15 Since when x=0, y=15; I know t      Log On


   

Question 48318: Given a table of x and F(x) values how would you find the quadratic equation that describes it?
x -4, -3, -2, -1, 0
y -1, 0, 3, 8, 15
Since when x=0, y=15; I know that the constant term in the quadratic equation must be 15. Also I know that one of the roots is -3 because when y=0, x=-3. Additionally I have found that the second difference beteen the "y" values is always +2. (pattern of change is 2). I however, do not know how to use this piece of information to solve the problem. I have to write the quadratic equation that describes the relationship. I know that the relationship is quadratic because the second difference is a constant term - 2. Please help. Thank you for your time and help.
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
well done with what you said and have done so far.

Right then, my method:

assuming it is a quadratic (and as you say, the second difference being constant implies a quadratic) then the equation will have the form of f(x) = +ax%5E2+%2B+bx+%2B+c+.

We know:
when x = 0 that y=15, so:
+f%28x%29+=+ax%5E2+%2B+bx+%2B+c+
+15+=+a%280%29%5E2+%2B+b%280%29+%2B+c+
--> c = 15

So we have +f%28x%29+=+ax%5E2+%2B+bx+%2B+15+

Knowing one of the roots is fine but it doesn't help us since there are many quadratic that will have that as a root. What we need to do now (seeing that we only have 2 unknowns to find --> a and b) is pick two of the coordinates and solve them simultaneously. Pick easier ones.

Using (-3,0) gives +0+=+a%28-3%29%5E2+-+3b+%2B+15+
+0+=+9a+-+3b+%2B+15+
--> 9a - 3b = -15

Using (-1,8) gives +8+=+a%28-1%29%5E2+-+b+%2B+15+
+8+=+a+-+b+%2B+15+
--> a - b = -7

So we have 2 equations now:
9a - 3b = -15
a - b = -7

scale up the second one:
9a - 3b = -15
3a - 3b = -21

and subtract them:
6a = 6
--> a = 1

and from a - b = -7 we get
1 - b = -7
--> b = 8

So the quadratic is +f%28x%29+=+x%5E2+%2B+8x+%2B+15+ which can be factorised to f(x) = (x+3)(x+5) meaning that its two roots are at -3 and -5, in agreement with the data quoted.

jon.