\( f(x)=\frac{x^{2}+3 x+4}{x-1} \) Choose the correct answer below. A. The \( x \)-intercept is \( (-3,0) \). There is a hole in its graph at \( x=1 \). C. The \( x \)-axis is its horizontal asymptote. D. The \( y \)-intercept is ( 0,1 ). The horizontal asymptote is \( y=1 \). F. The \( y \)-axis is its vertical asymptote. G. The graph has an oblique asymptote. H. The vertical asymptote is \( x=-1 \).

To analyze the function \( f(x) = \frac{x^{2} + 3x + 4}{x - 1} \), we will determine the \( x \)-intercept, \( y \)-intercept, vertical and horizontal asymptotes, and any holes in the graph. ### Step 1: Find the \( x \)-intercept The \( x \)-intercept occurs when \( f(x) = 0 \). This happens when the numerator is zero: \[ x^{2} + 3x + 4 = 0 \] We will solve this quadratic equation using the discriminant method: \[ D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \] Since the discriminant is negative, there are no real roots. Therefore, there is no \( x \)-intercept. ### Step 2: Find the \( y \)-intercept The \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = \frac{0^{2} + 3 \cdot 0 + 4}{0 - 1} = \frac{4}{-1} = -4 \] Thus, the \( y \)-intercept is \( (0, -4) \). ### Step 3: Find vertical asymptotes Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points): \[ x - 1 = 0 \implies x = 1 \] Thus, there is a vertical asymptote at \( x = 1 \). ### Step 4: Find horizontal asymptotes To find the horizontal asymptote, we compare the degrees of the numerator and denominator. The degree of the numerator (2) is greater than the degree of the denominator (1), so there is no horizontal asymptote. However, we can check for an oblique asymptote by performing polynomial long division. ### Step 5: Perform polynomial long division We will divide \( x^{2} + 3x + 4 \) by \( x - 1 \): 1. Divide the leading term: \( x^{2} \div x = x \). 2. Multiply \( x \) by \( x - 1 \): \( x(x - 1) = x^{2} - x \). 3. Subtract: \( (x^{2} + 3x + 4) - (x^{2} - x) = 4x + 4 \). 4. Divide the leading term: \( 4x \div x = 4 \). 5. Multiply \( 4 \) by \( x - 1 \): \( 4(x - 1) = 4x - 4 \). 6. Subtract: \( (4x + 4) - (4x - 4) = 8 \). Thus, the result of the division is: \[ f(x) = x + 4 + \frac{8}{x - 1} \] As \( x \to \infty \), the term \( \frac{8}{x - 1} \to 0 \), indicating that the oblique asymptote is \( y = x + 4 \). ### Summary of Findings - **\( x \)-intercept**: None - **\( y \)-intercept**: \( (0, -4) \) - **Vertical asymptote**: \( x = 1 \) - **Horizontal asymptote**: None (oblique asymptote \( y = x + 4 \)) ### Conclusion Now, let's evaluate the options: - A: Incorrect (no \( x \)-intercept, hole at \( x=1 \) is not applicable) - C: Incorrect (no horizontal asymptote) - D: Incorrect (the \( y \)-intercept is \( (0, -4) \), not \( (0, 1) \)) - F: Incorrect (the vertical asymptote is not at the \( y \)-axis) - G: Correct (there is an oblique asymptote) - H: Incorrect (the vertical asymptote is at \( x=1 \), not \( x=-1 \)) The correct answer is **G**: The graph has an oblique asymptote.