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How to solve for a quadratic equation when given three points?
For my math portfolio, I have been given the problem to algebraically solve the equation when given (-2,51), (1,-9), and (5-33).I plugged them into y=a(x)^2+b(x)+c. From there I got the equations;
51=4a-2b+c
-9=a+b+c
-33=25a+5b+c
If I eliminate -9=a+b+c then I get 60=3a-b and -24=24a+4b, but once I solve for a and b I get -22=a and -126=b. This is very off, and I don't know how to solve this. Could someone help me?
How do I eliminate it properly?
1 Answer
- ?Lv 77 years agoFavorite Answer
Your method is correct. You have your three equations right
51=4a-2b+c
-9=a+b+c
-33=25a+5b+c
You just got your elimination wrong, the correct answer is
a = 2, b = -18, c = 7
