Ответ:
Functions and Graphs
We need to match the functions to their corresponding graphs.
Functions:
- A) \( y = \frac{1}{2} x - 6 \)
- Б) \( y = x^2 - 8x + 11 \)
- B) \( y = -\frac{9}{x} \)
Graphs:
Graph 1) appears to be a parabola opening upwards, characteristic of a quadratic function. The vertex seems to be around (4, -5).
Graph 2) shows a hyperbola, which is characteristic of a rational function of the form \( y = \frac{k}{x} \).
Graph 3) is a straight line with a negative slope, characteristic of a linear function of the form \( y = mx + b \) where \( m < 0 \).
Matching:
- Function A) \( y = \frac{1}{2} x - 6 \) is a linear function. The slope is positive \( (\frac{1}{2}) \), and the y-intercept is -6. Graph 3) is a line, but the slope appears negative. Let's re-examine. Graph 3 shows a line with a POSITIVE slope passing through approximately (0, -6). This matches function A.
- Function Б) \( y = x^2 - 8x + 11 \) is a quadratic function. The graph of a quadratic function is a parabola. Graph 1) is a parabola. The vertex of \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). For this function, \( x = -\frac{-8}{2 \cdot 1} = 4 \). When \( x = 4 \), \( y = 4^2 - 8 \cdot 4 + 11 = 16 - 32 + 11 = -5 \). Graph 1) has its vertex at approximately (4, -5), so it matches function Б.
- Function B) \( y = -\frac{9}{x} \) is a rational function, a hyperbola. The graph of \( y = -\frac{k}{x} \) with \( k > 0 \) is in the second and fourth quadrants. Graph 2) shows such a hyperbola.
Table:
| A | Б | B |
|---|---|---|
| 3 | 1 | 2 |
Ответ: В таблице под каждой буквой укажите соответствующий номер: A - 3, Б - 1, B - 2.
