1. **State the problem:** We need to complete the table of values for the function $$f(x) = \frac{4}{x-2}$$ at the points $$x = -2, -1, 0, 1, 2, 3, 4$$. 2. **Recall the function and domain:** The function is a rational function with a denominator $$x-2$$. It is undefined where the denominator is zero, i.e., at $$x=2$$. 3. **Calculate each value:** - For $$x = -2$$: $$f(-2) = \frac{4}{-2 - 2} = \frac{4}{-4} = -1$$ - For $$x = -1$$: $$f(-1) = \frac{4}{-1 - 2} = \frac{4}{-3} = -\frac{4}{3}$$ - For $$x = 0$$: $$f(0) = \frac{4}{0 - 2} = \frac{4}{-2} = -2$$ - For $$x = 1$$: $$f(1) = \frac{4}{1 - 2} = \frac{4}{-1} = -4$$ - For $$x = 2$$: The denominator is zero, so $$f(2) = \text{N (undefined)}$$ - For $$x = 3$$: $$f(3) = \frac{4}{3 - 2} = \frac{4}{1} = 4$$ - For $$x = 4$$: $$f(4) = \frac{4}{4 - 2} = \frac{4}{2} = 2$$ 4. **Summary table:** | x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | |----|----|----|----|----|----|----|----| | f(x) | -1 | -\frac{4}{3} | -2 | -4 | N | 4 | 2 | 5. **Graph shape and description:** The function has a vertical asymptote at $$x=2$$ where it is undefined. For $$x<2$$, the function values are negative and approach negative infinity near $$x=2$$. For $$x>2$$, the function values are positive and decrease towards zero as $$x$$ increases. This completes the table and the understanding of the function behavior.