List the domain and the $$ x $$-and $$ y $$-intercepts of the following function. Graph the functio Be sure to label ahl the asymptotes. $$ f(x)=\frac{x^{2}-5 x+4}{x+1} $$

Question

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**Step 1. Factor and Simplify**

We start with the function  
$$
f(x)=\frac{x^{2}-5x+4}{x+1}.
$$  
Factor the numerator:
$$
x^{2}-5x+4=(x-1)(x-4).
$$  
So the function is written as
$$
f(x)=\frac{(x-1)(x-4)}{x+1}.
$$

---

**Step 2. Determine the Domain**

The only restriction occurs when the denominator is zero, i.e. when 
$$
x+1=0\quad\Longrightarrow\quad x=-1.
$$  
Thus, the domain of $$ f(x) $$ is
$$
\{x\in\mathbb{R} : x\neq -1\}.
$$

---

**Step 3. Find the $$x$$-intercepts**

The $$ x $$-intercepts occur when $$ f(x)=0 $$ (and the function is defined). Set the numerator equal to zero:
$$
(x-1)(x-4)=0.
$$
This gives:
$$
x-1=0\quad\Longrightarrow\quad x=1,
$$
$$
x-4=0\quad\Longrightarrow\quad x=4.
$$  
Thus, the $$ x $$-intercepts are at
$$
(1,0) \quad \text{and} \quad (4,0).
$$

---

**Step 4. Find the $$y$$-intercept**

To find the $$ y $$-intercept, set $$ x=0 $$:
$$
f(0)=\frac{0^{2}-5\cdot0+4}{0+1}=\frac{4}{1}=4.
$$
Thus, the $$ y $$-intercept is
$$
(0,4).
$$

---

**Step 5. Identify the Asymptotes**

1. **Vertical Asymptote**

The vertical asymptote occurs at the value excluded from the domain:
   $$
   x=-1.
   $$
   Since the numerator does not cancel with the denominator at $$ x=-1 $$, there is a vertical asymptote at
   $$
   x=-1.
   $$

2. **Oblique (Slant) Asymptote**

Because the degree of the numerator ($$2$$) is one more than the degree of the denominator ($$1$$), there is an oblique asymptote. To find it, perform polynomial long division of $$ x^2-5x+4 $$ by $$ x+1 $$:

- Divide $$ x^2 $$ by $$ x $$ to get $$ x $$. Multiply $$ x+1 $$ by $$ x $$ to get $$ x^2+x $$.
   - Subtract: 
     $$
     (x^2-5x+4) - (x^2+x) = -6x+4.
     $$
   - Divide $$ -6x $$ by $$ x $$ to get $$ -6 $$. Multiply $$ x+1 $$ by $$ -6 $$ to get $$ -6x-6 $$.
   - Subtract:
     $$
     (-6x+4)-(-6x-6)=10.
     $$
   
   The quotient is $$ x-6 $$ and the remainder is $$ 10 $$. Thus, we rewrite the function:
   $$
   f(x)=x-6+\frac{10}{x+1}.
   $$
   As $$ x \to \pm \infty $$, the fraction $$ \frac{10}{x+1} $$ approaches 0, and the graph approaches the line
   $$
   y=x-6.
   $$
   Therefore, the slant asymptote is
   $$
   y=x-6.
   $$

---

**Step 6. Summary of Results**

- **Domain:**  
  $$
  \{x\in\mathbb{R} : x\neq -1\}.
  $$

- **$$ x $$-intercepts:**  
  $$
  (1,0) \quad \text{and} \quad (4,0).
  $$

- **$$ y $$-intercept:**  
  $$
  (0,4).
  $$

- **Vertical Asymptote:**  
  $$
  x=-1.
  $$

- **Oblique (Slant) Asymptote:**  
  $$
  y=x-6.
  $$

---

**Step 7. Graphing the Function**

To graph $$ f(x) $$:

1. Plot and label the $$ x $$-intercepts at $$ (1,0) $$ and $$ (4,0) $$ and the $$ y $$-intercept at $$ (0,4) $$.

2. Draw the vertical asymptote as a dashed line at $$ x=-1 $$.

3. Draw the slant asymptote as a dashed line given by $$ y=x-6 $$.

4. Sketch the curve approaching the asymptotes:
   - For $$ x > -1 $$, the function behaves like $$ f(x)\approx x-6 $$ as $$ x $$ becomes large or very small.
   - Pay attention to the branch near $$ x=-1 $$, where the function diverges.

5. Ensure that the graph does not cross the vertical asymptote $$ x=-1 $$ and that it approaches the line $$ y=x-6 $$ as $$ x \to \pm\infty $$.

This completes the step-by-step process for listing the domain, intercepts, asymptotes, and graphing the function.
**Domain:** All real numbers except $$ x = -1 $$. **$$ x $$-intercepts:** $$ (1, 0) $$ and $$ (4, 0) $$. **$$ y $$-intercept:** $$ (0, 4) $$. **Asymptotes:** - **Vertical Asymptote:** $$ x = -1 $$. - **Oblique Asymptote:** $$ y = x - 6 $$. **Graph:** Plot the intercepts and asymptotes, then sketch the curve approaching these lines without crossing the vertical asymptote.

List the domain and the -and -intercepts of the following function. Graph the functio
Be sure to label ahl the asymptotes.

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List the domain and the -and -intercepts of the following function. Graph the functio Be sure to label ahl the asymptotes.