Hi,
So I missed my Pre-Ap Algebra II class, and I was wondering how I would be able to find a quadratic function that includes each set of Values.
So, the question that I was working on gave me three points, which are (1,-2),(2,-2), and (3,-4).
I did look in the textbook, and I attempted to try each. This is what I have so far.
*y=ax^2+bx=c
-2=A(-1)^2+b(1)+c
-2=a+b+c
*y=ax^2+bx=c
-2=a(2)^2+b(2)+c
-2=4a+2b+c
*y=ax^2+bx=c
-4=a(3)^2=b(3)=c
-4=9a+3b+c
So thats what I have so far, but in the example i noticed it said "use one of the methods in chap.3" and solve. Im not sure what to do after all of that from above. I just want to know which way is the easiest and will help me with my following homework problems. I was also wondering if there was any way to check it to make sure my answer will be right with this and the upcoming problems that I have to solve.
Thank you so much for your time and help
Your 3 equations in 3 unknowns are correct, up to this point.
You can use ELIMINATION (see below) to solve for the variables, and determine the quadratic equation.
You can also use MATRICES to solve for the variables, if you're familiar with this method.
a + b + c = - 2 -------- eq (i)
4a + 2b + c = - 2 -------- eq (ii)
9a + 3b + c = - 4 -------- eq (iii)
- 3a b = 0 ------- Subtracting eq (ii) from eq (i) ------ eq (iv)
- 5a - b = 2 ------- Subtracting eq (iii) from eq (ii) ------ eq (v)
2a = - 2 ----- Subtracting eq (v) from eq (iv)
a = , or
- 3(- 1) b = 0 ------- Substituting 1 for a in eq (iv)
3 b = 0
- b = - 3
b = , or
1 + (3) + c = - 2 ------- Substituting 1 for a, and 3 for b in eq (i)
1 + 3 + c = - 2
4 + c = - 2
c = - 2 - 2, or
With a being - 1, b being 3, and c being - 4, the equation: becomes:
Quadratic equation thus formed:
To check if this is the correct equation, just substitute each coordinate point into the equation, and each
equation should prove true.
You can do the check!!
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