4. for x≥0, the horizontal line y = 2 is an a...

# Explanation: ## Step1: Recall horizontal - asymptote definition A horizontal line \(y = L\) is a horizontal asymptote of the function \(y = f(x)\) if \(\lim_{x\rightarrow\infty}f(x)=L\) or \(\lim_{x\rightarrow-\infty}f(x)=L\). Given for \(x\geq0\), \(y = 2\) is a horizontal asymptote of \(y = f(x)\), so \(\lim_{x\rightarrow\infty}f(x)=2\). ## Step2: Analyze each option - Option (A): Just because \(y = 2\) is an asymptote doesn't mean \(f(0)=2\). The function value at \(x = 0\) has no relation to the asymptote value. - Option (B): The function can cross its horizontal asymptote. So \(f(x)\) can be equal to \(2\) for some \(x\geq0\). - Option (C): There is no information to suggest that \(f(2)\) is undefined. - Option (D): \(\lim_{x\rightarrow\infty}f(x)\neq\infty\) since \(y = 2\) is a horizontal asymptote. - Option (E): Since \(y = 2\) is a horizontal asymptote for \(x\geq0\), \(\lim_{x\rightarrow\infty}f(x)=2\). ## Step3: Recall inflection - point definition A point of inflection of the graph of \(y = f(x)\) occurs where \(f''(x)\) changes sign. Given \(f''(x)=x(x - a)(x - b)^2\). The roots of \(f''(x)\) are \(x = 0\), \(x=a\) and \(x = b\). Since \((x - b)^2\geq0\) for all \(x\), the sign of \(f''(x)\) changes at \(x = 0\) and \(x=a\) (because the factor \((x - b)^2\) does not change the sign at \(x = b\) as it is a squared - factor). # Answer: 4. E. \(\lim_{x\rightarrow\infty}f(x)=2\) 5. A. 0 and \(a\) only