1. Let's analyze the function $y=2\ln x$. - The domain of $\ln x$ is $x>0$, so the domain of $y=2\ln x$ is also $x>0$. - The range of $\ln x$ is $(-\infty, \infty)$, and multiplying by 2 does not change this, so the range is $(-\infty, \infty)$. - The vertical asymptote occurs where the function is undefined, which is at $x=0$. As $x \to 0^+$, $\ln x \to -\infty$, so $y \to -\infty$. - There is no horizontal asymptote because as $x \to \infty$, $\ln x \to \infty$ and so $y \to \infty$. 2. Now, analyze the function $y=2x10^x$. - The domain of $10^x$ is all real numbers, and multiplying by $2x$ does not restrict the domain, so the domain is $(-\infty, \infty)$. - To find the range, consider the behavior: - As $x \to -\infty$, $10^x \to 0$, so $y \to 2x \cdot 0 = 0$ from the negative side (since $x$ is negative), so $y \to 0^-$. - As $x \to \infty$, $10^x$ grows very fast, and since $x$ is positive, $y \to \infty$. - The function crosses zero at $x=0$ because $y=2 \cdot 0 \cdot 10^0=0$. - There are no vertical asymptotes because the function is defined for all real $x$. - There is a horizontal asymptote at $y=0$ as $x \to -\infty$. Final answers: - For $y=2\ln x$: - Domain: $x>0$ - Range: $(-\infty, \infty)$ - Vertical asymptote: $x=0$ - No horizontal asymptote - For $y=2x10^x$: - Domain: $(-\infty, \infty)$ - Range: $(-\infty, \infty)$ (since it goes to $0^-$ as $x \to -\infty$ and $\infty$ as $x \to \infty$) - Horizontal asymptote: $y=0$ as $x \to -\infty$ - No vertical asymptote