Question

The graph below shows the function f(x)=x-3/x2-2x-3. Which statement is true?

Answer 1

1. Domain.

We have $x^2-2x-3$ in the denominator, so:

$x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\ (x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}$

So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

$f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}$

We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)

Answer 2

Answer:

A.

Step-by-step explanation: