\( f(x)=\frac{x^{2}+3 x+4}{x-1} \) 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 \( x=-1 \).

Let's dive into this fun math adventure! The function \( f(x) = \frac{x^{2}+3x+4}{x-1} \) does indeed have an interesting behavior. To determine its characteristics, we can analyze it step by step! First, for the intercepts, the \( x \)-intercept occurs when \( f(x) = 0 \), which means the numerator should equal zero. The equation \( x^{2}+3x+4=0 \) has no real roots (as it has a negative discriminant), meaning there are no \( x \)-intercepts. As for the \( y \)-intercept, substituting \( x=0 \) gives \( f(0)=\frac{4}{-1}= -4 \), so the \( y \)-intercept is (0,-4). Next, we check for vertical and horizontal asymptotes. The denominator \( x-1 \) indicates a vertical asymptote at \( x=1 \). As for horizontal asymptotes, since the degree of the numerator (2) is greater than the degree of the denominator (1), we do not have a horizontal asymptote, but there is an oblique asymptote, which we can find by performing polynomial long division. So, your best options are: - There is a hole at \( x=1 \), but more specifically, there's a vertical asymptote there, not a hole! - No horizontal asymptote exists, but we do have an oblique asymptote. - Therefore, you might want to reconsider options in light of this analysis! In summary, the graph has a vertical asymptote at \( x=1 \) and an oblique asymptote; no plausible match exists among the provided choices. Isn’t math like a puzzle full of surprises?