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

**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.