V. CHALLENGE. Directions: Graph the rational function and find it's a) domain and range, b) \( x \) and \( y \)-intercepts, and c) horizontal and vertical asymptotes. (20 points) \( f(x)=\frac{x^{2}-3 x-4}{x+1} \)

To analyze the rational function \( f(x) = \frac{x^2 - 3x - 4}{x + 1} \), we will go through the steps to find the domain, range, intercepts, and asymptotes. ### Step 1: Domain The domain of a rational function is all real numbers except where the denominator is zero. Set the denominator equal to zero: \[ x + 1 = 0 \implies x = -1 \] Thus, the domain of \( f(x) \) is: \[ \text{Domain: } x \in \mathbb{R}, x \neq -1 \] ### Step 2: Range To find the range, we can analyze the behavior of the function. First, we can simplify the function by factoring the numerator: \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] Thus, we can rewrite the function as: \[ f(x) = \frac{(x - 4)(x + 1)}{x + 1} \] For \( x \neq -1 \), this simplifies to: \[ f(x) = x - 4 \] However, at \( x = -1 \), the function is undefined. The line \( y = x - 4 \) has a slope of 1 and a y-intercept of -4. The function approaches this line but is undefined at \( x = -1 \). To find the range, we note that as \( x \) approaches -1 from the left, \( f(x) \) approaches -5, and as \( x \) approaches -1 from the right, \( f(x) \) approaches -5 as well. Therefore, the function can take all real values except for \( y = -5 \). Thus, the range is: \[ \text{Range: } y \in \mathbb{R}, y \neq -5 \] ### Step 3: Intercepts **X-intercept:** Set \( f(x) = 0 \): \[ \frac{x^2 - 3x - 4}{x + 1} = 0 \implies x^2 - 3x - 4 = 0 \] Factoring gives: \[ (x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1 \] Since \( x = -1 \) is not in the domain, the x-intercept is: \[ \text{X-intercept: } (4, 0) \] **Y-intercept:** Set \( x = 0 \): \[ f(0) = \frac{0^2 - 3(0) - 4}{0 + 1} = \frac{-4}{1} = -4 \] Thus, the y-intercept is: \[ \text{Y-intercept: } (0, -4) \] ### Step 4: Asymptotes **Vertical Asymptote:** This occurs where the denominator is zero and the function is undefined. We already found that: \[ x + 1 = 0 \implies x = -1 \] Thus, the vertical asymptote is: \[ \text{Vertical Asymptote: } x = -1 \] **Horizontal Asymptote:** Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote, which is given by the simplified function: \[ y = x - 4 \] Thus, the oblique asymptote is: \[ \text{Oblique Asymptote: } y = x - 4 \] ### Summary - **Domain:** \( x \in \mathbb{R}, x \neq -1 \) - **Range:** \( y \in \mathbb{R}, y \neq -5 \) - **X-intercept:** \( (4, 0) \) - **Y-intercept:** \( (0, -4) \) - **Vertical Asymptote:** \( x = -1 \) - **Oblique Asymptote:** \( y = x - 4 \) ### Graphing To graph \( f(x) \), plot the intercepts, asymptotes, and the behavior of the function around the vertical asymptote. The function will approach the line \( y = x - 4 \) but will not touch or cross \( y = -5 \).