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

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**1. Domain**

The function is 
$$
f(x)=\frac{x^2-5x+4}{x+1}.
$$
The denominator $$ x+1=0 $$ when $$ x=-1 $$. Therefore, the domain is all real numbers except $$ x=-1 $$:
$$
\text{Domain: } \{ x\in\mathbb{R} : x\neq -1 \}.
$$

**2. $$ x $$-Intercepts**

The $$ x $$-intercepts occur when $$ f(x)=0 $$. Set the numerator equal to zero:
$$
x^2-5x+4=0.
$$
Factor the quadratic:
$$
x^2-5x+4=(x-1)(x-4)=0.
$$
Thus, the solutions are:
$$
x=1 \quad \text{and} \quad x=4.
$$
Since neither of these values makes the denominator zero, the $$ x $$-intercepts are:
$$
(1,0) \quad \text{and} \quad (4,0).
$$

**3. $$ y $$-Intercept**

The $$ y $$-intercept is found by evaluating $$ f(0) $$:
$$
f(0)=\frac{0^2-5(0)+4}{0+1}=\frac{4}{1}=4.
$$
Thus, the $$ y $$-intercept is:
$$
(0,4).
$$

**4. Asymptotes**

*Vertical Asymptote:*  
The vertical asymptote occurs where the denominator equals zero and the numerator is nonzero. Since
$$
x+1=0 \quad \text{when} \quad x=-1,
$$
and
$$
f(-1)=\text{undefined},
$$
there is a vertical asymptote at:
$$
x=-1.
$$

*Oblique (Slant) Asymptote:*  
Since the degree of the numerator (2) exceeds that of the denominator (1) by 1, an oblique asymptote exists. Perform polynomial division of $$ x^2-5x+4 $$ by $$ x+1 $$.

- **Divide:**  
  $$
  \frac{x^2}{x}=x.
  $$
  Multiply: $$ x(x+1)=x^2+x $$.

- **Subtract:**  
  $$
  (x^2-5x+4)-(x^2+x)=-6x+4.
  $$

- **Divide:**  
  $$
  \frac{-6x}{x}=-6.
  $$
  Multiply: $$ -6(x+1)=-6x-6 $$.

- **Subtract:**  
  $$
  (-6x+4)-(-6x-6)=10.
  $$

Thus, the division gives:
$$
f(x)=x-6+\frac{10}{x+1}.
$$
As $$ x\to \pm\infty $$, the term $$ \frac{10}{x+1} $$ approaches zero. Therefore, the oblique asymptote is:
$$
y=x-6.
$$

**5. Graphing the Function**

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

- **Plot the asymptotes:**  
  - Vertical asymptote: the line $$ x=-1 $$.  
  - Oblique asymptote: the line $$ y=x-6 $$.

- **Behavior near the asymptotes:**  
  - As $$ x $$ approaches $$-1$$ from the left and right, the function values will tend toward $$\pm\infty$$.  
  - For large positive or negative $$ x $$, the graph will get closer and closer to the line $$ y=x-6 $$.

- **Sketching the curve:**  
  - Draw a dashed vertical line at $$ x=-1 $$.  
  - Draw a dashed line with equation $$ y=x-6 $$ (a line with slope 1 that crosses the $$ y $$-axis at $$-6$$).  
  - Plot the intercept points $$(1,0)$$, $$(4,0)$$, and $$(0,4)$$.  
  - Sketch the curve using these key features, ensuring that the branches of the graph approach the asymptotes accordingly.

The final graph will show a hyperbola-like curve (modified by the linear asymptote) with a break at $$ x=-1 $$ and the intercepts as calculated.
**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:** The function has a vertical asymptote at $$ x = -1 $$ and an oblique asymptote at $$ y = x - 6 $$. It crosses the $$ x $$-axis at $$ (1,0) $$ and $$ (4,0) $$, and the $$ y $$-axis at $$ (0,4) $$. The graph approaches the asymptotes as $$ x $$ approaches $$-1$$ and extends towards $$ y = x - 6 $$ for large values of $$ x $$.

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

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