Question

даю 90 баллов,очень срочно![завдання 2]

Answer 1

Ответ:

$2) \ \ y=\dfrac{5x}{x-3}$

===============

ОДЗ:  $x-3\neq 0$

          $x\neq 3$

===============

$y_0=\dfrac{5*0}{0-3}\\\\\\y_0=\dfrac{0}{-3}\\\\\\y_0=0$

    $\downarrow$

$\boxed{(0; \ 0)}$

===========================================================

$4) \ \ y=\dfrac{2}{x-5}+4$

===============

ОДЗ:  $x-5\neq 0$

          $x\neq 5$

===============

$y_0=\dfrac{2}{0-5}+4 \qquad \qquad 0=\dfrac{2}{x_0-5}+4\\\\\\y_0=\dfrac{2}{-5}+4 \qquad \qquad \quad 0=2+4x_0-20\\\\\\y_0=3,6 \qquad \qquad \qquad \quad 18=4x_0\\\\\downarrow \qquad \downarrow \qquad \qquad \qquad \quad \ \ x_0=4,5\\\\\boxed{(0;\ 3,6)} \qquad \qquad \qquad \quad \downarrow \qquad \downarrow$

                                      $\boxed{(4,5; \ 0)}$

===========================================================

$6) \ \ y=x^2-2x-3\\\\{x_0}^{2}-2x_0-3=0 \\==================\\\\D = b^2-4ac\\\\a=1; \quad b = -2; \quad c=-3;\\\\D=(-2)^2-4 \times 1 \times (-3)\\\\D=16\\\\==================\\\\x_{1,2}=\dfrac{-b \pm \sqrt{D}}{2a}\\\\\\x_{1,2}=\dfrac{-(-2) \pm \sqrt{16}}{2 \times1}\\\\\\x_{1,2}=\dfrac{2 \pm 4}{2}\\\\\\x_1 = -1; \qquad x_2 = 3\\\\\downarrow \qquad \ \ \downarrow \qquad \ \downarrow \qquad \downarrow \qquad \\\\\boxed{(-1; \ 0)} \qquad \boxed{(3; \ 0)}$

$y_0 = 0^2-2 \times 0 - 3\\\\y_0 = 0-3\\\\y_0 = -3$

     $\downarrow$

$\boxed{(0; \ -3)}$