lập PT parabol y=x/2+bx+c a) Biết(P) đi qua A(1,6);B(-2,6);C(0,-4) b) Biết(P) đi qua A(-1,2);B(2,0);C(3,1) 25/09/2021 Bởi Sarah lập PT parabol y=x/2+bx+c a) Biết(P) đi qua A(1,6);B(-2,6);C(0,-4) b) Biết(P) đi qua A(-1,2);B(2,0);C(3,1)
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: \(\eqalign{ & y = a{x^2} + bx + c \cr & a)\,\,Di\,\,qua\,\,A\left( {1;6} \right) \Rightarrow a + b + c = 6 \cr & \,\,\,\,\,Di\,\,qua\,\,B\left( { – 2;6} \right) \Rightarrow 4a – 2b + c = 6 \cr & \,\,\,\,\,Di\,\,qua\,C\left( {0; – 4} \right) \Rightarrow c = – 4 \cr & \Rightarrow \left\{ \matrix{ a + b + c = 6 \hfill \cr 4a – 2b + c = 6 \hfill \cr c = – 4 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ a = 5 \hfill \cr b = 5 \hfill \cr c = – 4 \hfill \cr} \right. \cr & b)\,\,Di\,\,qua\,\,A\left( { – 1;2} \right) \Rightarrow a – b + c = 2 \cr & \,\,\,\,\,Di\,\,qua\,\,B\left( {2;0} \right) \Rightarrow 4a + 2b + c = 0 \cr & \,\,\,\,\,Di\,\,qua\,C\left( {3;1} \right) \Rightarrow 9a + 3b + c = 1 \cr & \Rightarrow \left\{ \matrix{ a – b + c = 2 \hfill \cr 4a + 2b + c = 0 \hfill \cr 9a + 3b + c = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ a = {5 \over {12}} \hfill \cr b = – {{13} \over {12}} \hfill \cr c = {1 \over 2} \hfill \cr} \right. \cr} \) Bình luận
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:
\(\eqalign{
& y = a{x^2} + bx + c \cr
& a)\,\,Di\,\,qua\,\,A\left( {1;6} \right) \Rightarrow a + b + c = 6 \cr
& \,\,\,\,\,Di\,\,qua\,\,B\left( { – 2;6} \right) \Rightarrow 4a – 2b + c = 6 \cr
& \,\,\,\,\,Di\,\,qua\,C\left( {0; – 4} \right) \Rightarrow c = – 4 \cr
& \Rightarrow \left\{ \matrix{
a + b + c = 6 \hfill \cr
4a – 2b + c = 6 \hfill \cr
c = – 4 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
a = 5 \hfill \cr
b = 5 \hfill \cr
c = – 4 \hfill \cr} \right. \cr
& b)\,\,Di\,\,qua\,\,A\left( { – 1;2} \right) \Rightarrow a – b + c = 2 \cr
& \,\,\,\,\,Di\,\,qua\,\,B\left( {2;0} \right) \Rightarrow 4a + 2b + c = 0 \cr
& \,\,\,\,\,Di\,\,qua\,C\left( {3;1} \right) \Rightarrow 9a + 3b + c = 1 \cr
& \Rightarrow \left\{ \matrix{
a – b + c = 2 \hfill \cr
4a + 2b + c = 0 \hfill \cr
9a + 3b + c = 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
a = {5 \over {12}} \hfill \cr
b = – {{13} \over {12}} \hfill \cr
c = {1 \over 2} \hfill \cr} \right. \cr} \)
Thay lần lượt toạ độ các điểm đi qua vào $y=ax^2+bx+c$. Giải hệ ba ẩn tìm a, b, c.