Заполним таблицу, подставляя значения \( x \) в выражения для \( y = f(x) \) и вычисляя значения функции.
| № | \( y=f(x) \) | \( f(2) \) | \( f(0) \) | \( f(-1) \) |
| 1) | \( \frac{3}{x+2} \) | \( \frac{3}{2+2} = \frac{3}{4} \) | \( \frac{3}{0+2} = \frac{3}{2} \) | \( \frac{3}{-1+2} = 3 \) |
| 2) | \( \frac{4}{x-3} \) | \( \frac{4}{2-3} = -4 \) | \( \frac{4}{0-3} = -\frac{4}{3} \) | \( \frac{4}{-1-3} = -1 \) |
| 3) | \( \frac{6}{x-5} \) | \( \frac{6}{2-5} = -2 \) | \( \frac{6}{0-5} = -\frac{6}{5} \) | \( \frac{6}{-1-5} = -1 \) |
| 4) | \( \frac{-12}{x+4} \) | \( \frac{-12}{2+4} = -2 \) | \( \frac{-12}{0+4} = -3 \) | \( \frac{-12}{-1+4} = -4 \) |
| 5) | \( \frac{-9}{1-2x} \) | \( \frac{-9}{1-2(2)} = \frac{-9}{1-4} = 3 \) | \( \frac{-9}{1-2(0)} = -9 \) | \( \frac{-9}{1-2(-1)} = \frac{-9}{1+2} = -3 \) |
| 6) | \( y=x^2 \) | \( 2^2 = 4 \) | \( 0^2 = 0 \) | \( (-1)^2 = 1 \) |
| 7) | \( y=x^2+3 \) | \( 2^2+3 = 7 \) | \( 0^2+3 = 3 \) | \( (-1)^2+3 = 4 \) |
| 8) | \( y=30-x^2 \) | \( 30-2^2 = 26 \) | \( 30-0^2 = 30 \) | \( 30-(-1)^2 = 29 \) |
| 9) | \( y=25-x^2 \) | \( 25-2^2 = 21 \) | \( 25-0^2 = 25 \) | \( 25-(-1)^2 = 24 \) |
| 10) | \( y=-x^2 \) | \( -(2^2) = -4 \) | \( -(0^2) = 0 \) | \( -(-1)^2 = -1 \) |
| 11) | \( y=-x^2+4 \) | \( -(2^2)+4 = 0 \) | \( -(0^2)+4 = 4 \) | \( -(-1)^2+4 = 3 \) |
| 12) | \( y=-x^2-9 \) | \( -(2^2)-9 = -13 \) | \( -(0^2)-9 = -9 \) | \( -(-1)^2-9 = -10 \) |
| 13) | \( y=\sqrt{x+1} \) | \( \sqrt{2+1} = \sqrt{3} \) | \( \sqrt{0+1} = 1 \) | \( \sqrt{-1+1} = 0 \) |
| 14) | \( y=-\sqrt{x+2} \) | \( -\sqrt{2+2} = -2 \) | \( -\sqrt{0+2} = -\sqrt{2} \) | \( -\sqrt{-1+2} = -1 \) |
Ответ: Таблица заполнена согласно расчетам.