Cho hàm số \(y = - {x^2}\). Tìm tất cả các điểm trên (P) có tung độ \(- 3, - \dfrac{3}{2}.\)
A.A.
\(\left( {\sqrt 3 ; - 3} \right);\,\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right);\left( { - \dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
\(\left( {\sqrt 3 ; - 3} \right);\,\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right);\left( { - \dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
B.B.
\(\,\left( { \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
\(\,\left( { \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
C.C.
\(\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
\(\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
D.D.
\(\left( {\sqrt 3 ; - 3} \right);\,\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
\(\left( {\sqrt 3 ; - 3} \right);\,\left( { - \sqrt 3 ; - 3}\right); \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\)
Đáp án và lời giải
Đáp án:A
Lời giải:
\(\begin{array}{l}y = - {x^2}\\y = - 3 \Rightarrow - {x^2} = - 3 \Leftrightarrow x = \pm \sqrt 3 \\\Rightarrow \,\left( {\sqrt 3 ; - 3} \right);\,\left( { - \sqrt 3 ; - 3} \right)\\y = - \dfrac{3}{2} \Rightarrow - {x^2} = - \dfrac{3}{2} \Leftrightarrow x = \pm \dfrac{{\sqrt 6 }}{2}\\ \Rightarrow \left( {\dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right);\left( { - \dfrac{{\sqrt 6 }}{2}; - \dfrac{3}{2}} \right)\end{array}\)