To analyze how the graph $f(x) = \frac{x - 1}{x - 2}$ would look with a hole at $(-1, 3)$, follow these steps:

Step 1: Understanding the original graph

1. Domain: $f(x)$ is undefined at $x = 2$ because the denominator becomes 0. This creates a vertical asymptote at $x = 2$.
2. Horizontal Asymptote: For large values of $x$, the function behaves like $\frac{x}{x} = 1$, so there is a horizontal asymptote at $y = 1$.
3. Behavior near $x = 1$: The numerator becomes 0 at $x = 1$, so the function has an x-intercept at $x = 1$.

Step 2: Adjusting for the hole at $(-1, 3)$

To introduce a hole at $x = -1$, the function must be modified to reflect that $(-1, 3)$ is a removable discontinuity. A possible function is:

$f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 1)}$

Here, the factor $(x + 1)$ in the numerator and denominator cancels out, leaving $f(x) = \frac{x - 1}{x - 2}$ for all $x \neq -1$. At $x = -1$, $f(x)$ is undefined, creating a hole at $(-1, 3)$.

Step 3: Verifying the hole's value

Substitute $x = -1$ into the simplified form $f(x) = \frac{x - 1}{x - 2}$:

$f(-1) = \frac{-1 - 1}{-1 - 2} = \frac{-2}{-3} = \frac{2}{3}.$

Therefore, the point $(-1, 3)$ is where the hole would occur, as $f(-1)$ is not defined.

Step 4: Graphing details

• Key features:

  1. Vertical asymptote: $x = 2$.
  2. Horizontal asymptote: $y = 1$.
  3. X-intercept: $x = 1$.
  4. A hole at $(-1, 3)$.

• Shape: The graph approaches the asymptotes and passes through the x-intercept, but skips the hole.

Would you like me to draw this graph or provide further steps?