Question

найдите точки пересечения парабулы y=x в квадрате и прямой; а) x=0 б)y=2x

Answer 1

$\displaystyle \left \{ {{y=x^2} \atop {x=0}} \right.\quad\rightarrow\quad y=0,\quad \quad \boxed{(0;0)}\\\\\\\left \{ {{y=x^2} \atop {y=2x}} \right. \quad\rightarrow\quad x^2=2x\\\\\\x^2-2x=0\\\\x(x-2)=0\\\\\left[\begin{array}{ccc}x=0\\x=2\end{array}\right\quad\rightarrow\quad \left[\begin{array}{ccc}y(0)=0^2=0\\y(2)=2^2=4\end{array}\right\quad\rightarrow\quad\boxed{\left[\begin{array}{ccc}(0;0)\\(2;4)\end{array}\right }$