Функция задана формулой $y = -\frac{1}{6}x + 2$. Найдите значение y, если:
1) значения функции для значения аргумента равных 12; 6; −6; 0; 1; 2; −4; −3;
2) значения аргумента, при котором значение функции равно 4; 3; 0; −1.
при x = 12:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * 12 + 2 = -2 + 2 = 0$;
при x = 6:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * 6 + 2 = -1 + 2 = 1$;
при x = −6:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * -6 + 2 = 1 + 2 = 3$;
при x = 0:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * 0 + 2 = 0 + 2 = 2$;
при x = 1:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * 1 + 2 = -\frac{1}{6} + 2 = 1\frac{5}{6}$;
при x = 2:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * 2 + 2 = -\frac{1}{3} + 2 = 1\frac{2}{3}$;
при x = −4:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * -4 + 2 = \frac{2}{3} + 2 = 2\frac{2}{3}$;
при x = −3:
$y = -\frac{1}{6}x + 2 = -\frac{1}{6} * -3 + 2 = \frac{1}{2} + 2 = 2\frac{1}{2}$.
при y = 4:
$y = -\frac{1}{6}x + 2$
$4 = -\frac{1}{6}x + 2$
$\frac{1}{6}x = 2 - 4$
$\frac{1}{6}x = -2$
$x = -2 : \frac{1}{6}$
x = −2 * 6
x = −12;
при y = 3:
$y = -\frac{1}{6}x + 2$
$3 = -\frac{1}{6}x + 2$
$\frac{1}{6}x = 2 - 3$
$\frac{1}{6}x = -1$
$x = -1 : \frac{1}{6}$
x = −1 * 6
x = −6;
при y = 0:
$y = -\frac{1}{6}x + 2$
$0 = -\frac{1}{6}x + 2$
$\frac{1}{6}x = 2$
$x = 2 : \frac{1}{6}$
x = 2 * 6
x = 12;
при y = −1:
$y = -\frac{1}{6}x + 2$
$-1 = -\frac{1}{6}x + 2$
$\frac{1}{6}x = 2 + 1$
$\frac{1}{6}x = 3$
$x = 3 : \frac{1}{6}$
x = 3 * 6
x = 18.
Пожаулйста, оцените решение
