Summarize all pertinent information obtained by applying the graphing strategy and sketch the graph of \( y=f(x) \). \[ f(x)=\frac{x^{2}+10 x+24}{x^{2}+8 x+16} \] Find any horizontal asymptotes of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one horizontal asymptote, \( \square \) (Type an equation.) B. The function has two horizontal asymptotes. The top asymptote is \( \square \) and the bottom asymptote is \( \square \) . (Type equations.) C. There are no horizontal asymptotes. Find any vertical asymptotes of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, \( \square \) (Type an equation.) B. The function has two vertical asymptotes. The leftmost asymptote is \( \square \) and the rightmost asymptote is \( \square \) . (Type equations.) C. There are no vertical asymptotes.
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To analyze the function \( f(x) = \frac{x^{2}+10x+24}{x^{2}+8x+16} \), we first factor both the numerator and denominator. The numerator factors to \( (x+6)(x+4) \) and the denominator to \( (x+4)(x+4) = (x+4)^2 \). This indicates that there is a removable discontinuity at \( x = -4 \) but no vertical asymptote there. The degree of the numerator and the denominator are the same, so we can find the horizontal asymptote by taking the leading coefficients. Thus, the horizontal asymptote is \( y = 1 \). Horizontal asymptote choice: A. The function has one horizontal asymptote, \( y = 1 \). For vertical asymptotes, we need to check when the denominator is \( 0 \). The denominator \( (x + 4)^2 \) is \( 0 \) when \( x = -4 \), but since this is where the function is undefined (removable discontinuity) rather than an asymptote, we conclude that there are no vertical asymptotes. Vertical asymptote choice: C. There are no vertical asymptotes.
