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Find real numbers a, b, and c so that the graph of the function y=ax^2+bx+c contains the points (-1,3), (3,8), and (0,2).

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konrad509
[tex]3=a\cdot(-1)^2+b\cdot(-1)+c\\ 8=a\cdot3^2+b\cdot3+c\\ 2=a\cdot0^2+b\cdot0+c\\\\ 3=a-b+c\\ 8=9a+3b+c\\ 2=c\\\\ 3=a-b+2\\ 8=9a+3b+2\\\\ a=b+1\\ 9a+3b=6\\\\ 9(b+1)+3b=6\\ 9b+9+3b=6\\ 12b=-3\\ b=-\frac{3}{12}=-\frac{1}{4}\\\\ a=-\frac{1}{4}+1\\ a=\frac{3}{4}\\\\ \boxed{y=\frac{3}{4}x^2-\frac{1}{4}x+2}[/tex]
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