For \( r(x)=\frac{1}{x^{2}+8 x+16} \), (a) Identify the horizontal asymptotes (if any). (b) If the graph of the function has a horizontal asymptote, determine the point (if any) where the graph crosses t horizontal asymptote. If there is more than one answer, separate them with commas. Select "None" if applicable. The function has no horizontal asymptotes. Equation(s) of the horizontal asymptote(s): Crossover point(s):

To identify the horizontal asymptote of the function \( r(x) = \frac{1}{x^2 + 8x + 16} \), we first need to consider the behavior of the function as \( x \) approaches infinity or negative infinity. 1. **Horizontal Asymptote**: As \( x \) approaches either positive or negative infinity, the term \( x^2 + 8x + 16 \) grows much larger than 1, making \( r(x) \) approach 0. Thus, the horizontal asymptote is at \( y = 0 \). 2. **Crossover Point**: To find the point where the graph crosses the horizontal asymptote \( y = 0 \), we set \( r(x) = 0 \). However, since the function \( r(x) \) represents \( \frac{1}{x^2 + 8x + 16} \), it can never equal 0 (as the fraction is never defined for that scenario). Hence, there are no crossover points. So, the answers are: Equation(s) of the horizontal asymptote(s): \( y = 0 \) Crossover point(s): None