Đáp án:
\(a)\,\,\left\{ \matrix{
a = 5 \hfill \cr
b = 5 \hfill \cr
c = - 4 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\left\{ \matrix{
a = {5 \over {12}} \hfill \cr
b = - {{13} \over {12}} \hfill \cr
c = {1 \over 2} \hfill \cr} \right.\)
Giải thích các bước giải:
\(\eqalign{
& y = a{x^2} + bx + c \cr
& a)\,\,Di\,\,qua\,\,A\left( {1;6} \right) \Rightarrow a + b + c = 6 \cr
& \,\,\,\,\,Di\,\,qua\,\,B\left( { - 2;6} \right) \Rightarrow 4a - 2b + c = 6 \cr
& \,\,\,\,\,Di\,\,qua\,C\left( {0; - 4} \right) \Rightarrow c = - 4 \cr
& \Rightarrow \left\{ \matrix{
a + b + c = 6 \hfill \cr
4a - 2b + c = 6 \hfill \cr
c = - 4 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
a = 5 \hfill \cr
b = 5 \hfill \cr
c = - 4 \hfill \cr} \right. \cr
& b)\,\,Di\,\,qua\,\,A\left( { - 1;2} \right) \Rightarrow a - b + c = 2 \cr
& \,\,\,\,\,Di\,\,qua\,\,B\left( {2;0} \right) \Rightarrow 4a + 2b + c = 0 \cr
& \,\,\,\,\,Di\,\,qua\,C\left( {3;1} \right) \Rightarrow 9a + 3b + c = 1 \cr
& \Rightarrow \left\{ \matrix{
a - b + c = 2 \hfill \cr
4a + 2b + c = 0 \hfill \cr
9a + 3b + c = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
a = {5 \over {12}} \hfill \cr
b = - {{13} \over {12}} \hfill \cr
c = {1 \over 2} \hfill \cr} \right. \cr} \)
