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The graph below shows the function f(x)=x-3/x2-2x-3. Which statement is true?
There is a hole at x = 3 and an asymptote at x = –1.
There is an asymptote at x = –1 and no hole.
There is a hole at x = 3 and no asymptote.
There is an asymptote at x = 3 and a hole at x = –1.

Respuesta :

WojtekR
1. Domain.

We have [tex]x^2-2x-3 [/tex] in the denominator, so:

[tex]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}[/tex]

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

2. Asymptotes:

[tex]f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}[/tex]

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

The given function has a hole at (x = 3) and an asymptote at (x = -1) and this can be determined by using the given data.

Given :

[tex]\rm f(x) = \dfrac{x-3}{x^2-2x-3}[/tex]

The following steps can be used in order to determine the true statement about the given function:

Step 1 - Write the given function.

[tex]\rm f(x) = \dfrac{x-3}{x^2-2x-3}[/tex]

Step 2 - The denominator of the following function never be zero.

[tex]x^2-2x-3\neq 0[/tex]

[tex]x^2-3x+x-3\neq 0[/tex]

[tex]x(x-3)+1(x-3)\neq 0[/tex]

[tex](x-3)(x+1)\neq 0[/tex]

Step 3 - Now, simplify the given function.

[tex]\rm f(x) = \dfrac{x-3}{(x-3)(x+1)}[/tex]

[tex]\rm f(x) = \dfrac{1}{x+1}[/tex]

Step 4 - So, the asymptote is given below:

x + 1 = 0

x = -1

Therefore, the correct option is A).

For more information, refer to the link given below:

https://brainly.com/question/15324782

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