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МатематикаВопрос:
Fill in the missing values in the table, given the functions y = f(x) and specific values for f(1), f(0), and f(-2).
Ответ:
Table Completion
| y = f(x) | f(1) | f(0) | f(-2) |
| 1) y = x + 8 | f(1) = 1 + 8 = 9 | f(0) = 0 + 8 = 8 | f(-2) = -2 + 8 = 6 |
| 2) y = x - 7 | f(1) = 1 - 7 = -6 | f(0) = 0 - 7 = -7 | f(-2) = -2 - 7 = -9 |
| 3) y = 2x + 1 | f(1) = 2(1) + 1 = 3 | f(0) = 2(0) + 1 = 1 | f(-2) = 2(-2) + 1 = -3 |
| 4) y = 10 - 5x | f(1) = 10 - 5(1) = 5 | f(0) = 10 - 5(0) = 10 | f(-2) = 10 - 5(-2) = 20 |
| 5) y = 12 - 3x | f(1) = 12 - 3(1) = 9 | f(0) = 12 - 3(0) = 12 | f(-2) = 12 - 3(-2) = 18 |
| 6) \( y = \frac{1}{2}x + 7 \) | \( f(1) = \frac{1}{2}(1) + 7 = 7.5 \) | \( f(0) = \frac{1}{2}(0) + 7 = 7 \) | \( f(-2) = \frac{1}{2}(-2) + 7 = 6 \) |
| 7) \( y = 3 - \frac{1}{2}x \) | \( f(1) = 3 - \frac{1}{2}(1) = 2.5 \) | \( f(0) = 3 - \frac{1}{2}(0) = 3 \) | \( f(-2) = 3 - \frac{1}{2}(-2) = 4 \) |
| 8) \( y = \frac{x}{2} \) | \( f(1) = \frac{1}{2} \) | \( f(0) = 0 \) | \( f(-2) = \frac{-2}{2} = -1 \) |
| 9) \( y = \frac{x+1}{2} \) | \( f(1) = \frac{1+1}{2} = 1 \) | \( f(0) = \frac{0+1}{2} = 0.5 \) | \( f(-2) = \frac{-2+1}{2} = -0.5 \) |
| 10) \( y = \frac{2-x}{10} \) | \( f(1) = \frac{2-1}{10} = 0.1 \) | \( f(0) = \frac{2-0}{10} = 0.2 \) | \( f(-2) = \frac{2-(-2)}{10} = 0.4 \) |
| 11) \( y = -\frac{x}{10} - 1 \) | \( f(1) = -\frac{1}{10} - 1 = -1.1 \) | \( f(0) = -\frac{0}{10} - 1 = -1 \) | \( f(-2) = -\frac{-2}{10} - 1 = -0.8 \) |
| 12) \( y = -\frac{x}{2} - 0.5 \) | \( f(1) = -\frac{1}{2} - 0.5 = -1 \) | \( f(0) = -\frac{0}{2} - 0.5 = -0.5 \) | \( f(-2) = -\frac{-2}{2} - 0.5 = 0.5 \) |
