$$ \mid \leftarrow $$ List the domain and the $$ x $$ - and $$ y $$-intercepts of the following function. Graph the function. Be sure to label all the asymptotes. $$ f(x)=\frac{x-1}{x^{2}-2 x-3} $$ What is the domain of the function? $$ (-\infty,-1) \cup(-1,3) \cup(3, \infty) $$ (Type your answer in interval notation.) Find the $$ x $$ - and $$ y $$-intercepts. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. (Type an ordered pair, using integers or fractions. Use a comma to separate answers as needed.) A. There is no $$ y $$-intercept. The $$ x $$-intercept(s) is/are $$ \square $$. B. There is no $$ x $$-intercept. The $$ y $$-intercept is $$ \square $$. The $$ x $$-intercept(s) is/are $$ \square $$. The $$ y $$-intercept is $$ \square $$. D. There are no $$ x $$ - and $$ y $$-intercepts.

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UpStudy AI · Accepted Answer

**Domain:**  
The function is  
$$
f(x)=\frac{x-1}{x^2-2x-3}
$$  
Factor the denominator:  
$$
x^2-2x-3=(x-3)(x+1)
$$  
The denominator is zero when $$x=3$$ or $$x=-1$$. Thus, the domain is all real numbers except $$x=-1$$ and $$x=3$$. In interval notation, the domain is  
$$
(-\infty,-1) \cup (-1,3) \cup (3,\infty)
$$

**Intercepts:**

1. **$$x$$-intercept:**  
Set $$f(x)=0$$. This occurs when the numerator is zero (provided the denominator is not zero).  
$$
x-1=0 \quad \Longrightarrow \quad x=1
$$  
Since $$1$$ is in the domain (the denominator is not zero at $$x=1$$), the $$x$$-intercept is  
$$
(1,0)
$$

2. **$$y$$-intercept:**  
Substitute $$x=0$$ into the function:  
$$
f(0)=\frac{0-1}{0^2-2\cdot0-3}=\frac{-1}{-3}=\frac{1}{3}
$$  
Thus, the $$y$$-intercept is  
$$
\left(0,\frac{1}{3}\right)
$$

**Asymptotes:**

1. **Vertical asymptotes:**  
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). As determined earlier, this happens at  
$$
x=-1 \quad \text{and} \quad x=3
$$  
Thus, the vertical asymptotes are the lines  
$$
x=-1 \quad \text{and} \quad x=3
$$

2. **Horizontal asymptote:**  
Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is  
$$
y=0
$$

**Summary of Answers:**

- **Domain:** $$(-\infty,-1) \cup (-1,3) \cup (3,\infty)$$  
- **$$x$$-intercept:** $$(1,0)$$  
- **$$y$$-intercept:** $$\left(0,\frac{1}{3}\right)$$  
- **Vertical asymptotes:** $$x=-1$$ and $$x=3$$  
- **Horizontal asymptote:** $$y=0$$

**Graphing the Function:**

- Plot the $$x$$-intercept $$(1,0)$$ and $$y$$-intercept $$\left(0,\frac{1}{3}\right)$$.  
- Draw vertical dashed lines at $$x=-1$$ and $$x=3$$ to indicate the vertical asymptotes.  
- Draw a horizontal dashed line at $$y=0$$ for the horizontal asymptote.  
- Sketch the curve approaching the asymptotes appropriately, noting that the function graph is divided by the vertical asymptotes into separate regions.

Choice A is correct with the following intercepts:  
- There is a $$y$$-intercept at $$\left(0,\frac{1}{3}\right)$$ and an $$x$$-intercept at $$(1,0)$$.
**Domain:** $$ (-\infty,-1) \cup (-1,3) \cup (3,\infty) $$ **Intercepts:** - $$x$$-intercept: $$(1,0)$$ - $$y$$-intercept: $$\left(0,\frac{1}{3}\right)$$ **Asymptotes:** - Vertical asymptotes at $$x=-1$$ and $$x=3$$ - Horizontal asymptote at $$y=0$$ **Summary:** - Domain: All real numbers except $$x=-1$$ and $$x=3$$. - $$x$$-intercept: $$(1,0)$$ - $$y$$-intercept: $$\left(0,\frac{1}{3}\right)$$ - Vertical asymptotes: $$x=-1$$ and $$x=3$$ - Horizontal asymptote: $$y=0$$

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