SOLUTION: Find a quadratic function in standard form for each set of points.
(0, 3), (1, –4), (2, –9)
(0 , –4), (1, 0), (2, 2)
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Question 178109: Find a quadratic function in standard form for each set of points.
(0, 3), (1, –4), (2, –9)
(0 , –4), (1, 0), (2, 2)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If a quadratic function  = ax^2 + bx + c)
has a graph containing a point )
then 
.
So, since contains the point )
we know that ^2 + b(0) + c = 3)
which is to say 
.
Similarly, since contains the point )
we know that ^2 + b(1) + c = -4)
. But since we already know that , we can say 
which is to say 
.
Continuing the process, the third point, )
, will result in: ^2 + b(2) + 3 = -9)
which is to say 
.
Now:
and
form a system of two linear equations in two variables. Given that the coefficents on both variables in the first equation is 1, this system lends itself well to solution by substitution.
Substituting this expression for a into the second equation yields:
 + 2b = -12 \text { } \Rightarrow \text{ }\math -4b -28 + 2b = -12 \text { } \Rightarrow \text{ }\math -2b = 16 \text { } \Rightarrow \text{ }\math b = -8)
Now that we know 
we can substitute this value into  - 7 = 1)
Now that we have values for a, b, and c, we can write the quadratic function specifically:
 = ax^2 + bx + c = x^2 - 8x + 3)
Do the other problem the same way.
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