Question

Найдите точки пересечения графиков функция, построив их: 1) у= 3х; у=х-1; у=-2х; у=х+2.​

Answer 1

Ответ:

Чтобы найти точки пересечения графиков функций , надо решить совместно  уравнения прямых .

$\bf 1)\ \left\{\begin{array}{l}\bf y=3x\\\bf y=x-1\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x=x-1\\\bf y=x-1\end{array}\right\ \ \left\{\begin{array}{l}\bf 2x=-1\\\bf y=x-1\end{array}\right\ \ \left\{\begin{array}{l}\bf x=-0,5\\\bf y=-1,5\end{array}\right$  

$\bf 2)\ \left\{\begin{array}{l}\bf y=3x\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x=-2x\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf 5x=0\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf x=0\\\bf y=0\end{array}\right$  

$\bf 3)\ \left\{\begin{array}{l}\bf y=3x\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x=x+2\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf 2x=2\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf x=1\\\bf y=3\end{array}\right$  

$\bf 4)\ \left\{\begin{array}{l}\bf y=x-1\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf x-1=-2x\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x=1\\\bf y=-2x\end{array}\right\ \ \left\{\begin{array}{l}\bf x=\dfrac{1}{3}\\\bf y=-\dfrac{2}{3}\end{array}\right$  

$\bf 5)\ \left\{\begin{array}{l}\bf y=x-1\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf x-1=x+2\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf -1=2\ neverno\\\bf y=x+2\end{array}\right$  

Прямые не пересекаются .

$\bf 6)\ \left\{\begin{array}{l}\bf y=-2x\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf -2x=x+2\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf 3x=-2\\\bf y=x+2\end{array}\right\ \ \left\{\begin{array}{l}\bf x=-\dfrac{2}{3}\\\bf y=\dfrac{4}{3}\end{array}\right$