Question 12588: I'm supposed to find a family of quadratic equations passing through the points (1,0) and (-1,-2), using the general form y=Ax^2+Bx+C. So far, here's what I got:
1) Plugged in both points to the general form, giving me 2 distinct equations:
0=A+B+C
0=A-B+C+2
2) Made them equal to each other, giving me B=1
3) Subbed B=1 into either equation to get A=-C-1 and C=-A-1
Now this is the point where I get stuck, since if I use both B=1 and A=-C-1 to get the final two variables, I end up with nothing. Please help, and thanks so much!
-Bing
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! You are correct. Remember that you are trying to find a family of parabolas, which means that there are perhaps an infinite number of such parabolas. This is what happens when you have three unknowns and only two conditions (i.e., equations) to satisfy.
Both equations come down to A + C = -1. So, let B=1, and solve for either C or A in terms of the other. Say, C= -A -1.
The equation is the family of parabolas:
 , where A represents any non-zero number. Of course, if A = 0, then it is NOT a parabola.
Any other value of A should give you points that include (1,0) and (-1,-2).
For examples, let A = 1:  Do both points satisfy this? If x =1, then y= 0, and if x=-1, then y =-2.
Let A = 2: (((y = 2x^2 +x -3}}}
Let A = -1: 
R^2 at SCC
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