Question

the graph below shows the function \\(f(x) = \frac{x-3}{x^2-2x-3}\\)

which statement is true?

there is a hole at \\(x = 3\\) and an asymptote at \\(x = -1\\).

Explanation:

Factor the denominator

Using the Polynomial Factoring knowledge point

$$ x^2 - 2x - 3 = (x - 3)(x + 1) $$

Simplify the rational function

Using the Rational Functions knowledge point

$$ f(x) = \frac{x - 3}{(x - 3)(x + 1)} = \frac{1}{x + 1}, \quad x eq 3 $$

Identify the hole

Using the Holes in Rational Graphs knowledge point

$$ \text{Since } x - 3 \text{ cancels, there is a hole at } x = 3. $$

Identify the vertical asymptote

Using the Vertical Asymptotes knowledge point

$$ \text{Since } x + 1 = 0 \text{ remains in the denominator, there is a vertical asymptote at } x = -1. $$

Answer:

• (A) There is a hole at \(x = 3\) and an asymptote at \(x = -1\). (Correct answer)