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nataliyantonov
@nataliyantonov
August 2022
1
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Помогите пожалуйста!!!
найдите точки пересечения графиков функций у=-х-1 и у=-(х+2)^2+3
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ShirokovP
Verified answer
- x - 1 = - (x + 2)^2 +3
- x - 1 = - (x^2 + 4x + 4) + 3
- x - 1 = - x^2 - 4x - 4 + 3
- x
- 1
= - x^2 - 4x
- 1
x^2 + 4x - x = 0
x^2 + 3x = 0
x (x + 3) = 0
x = 0 ==> y = - 0 - 1 = - 1
x = - 3 ==> y = 3 - 1 = 2
ОТВЕТ
( - 3; 2) ; (0; - 1)
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Answers & Comments
Verified answer
- x - 1 = - (x + 2)^2 +3- x - 1 = - (x^2 + 4x + 4) + 3
- x - 1 = - x^2 - 4x - 4 + 3
- x - 1 = - x^2 - 4x - 1
x^2 + 4x - x = 0
x^2 + 3x = 0
x (x + 3) = 0
x = 0 ==> y = - 0 - 1 = - 1
x = - 3 ==> y = 3 - 1 = 2
ОТВЕТ
( - 3; 2) ; (0; - 1)