1. **State the problem:** We need to analyze the function $$y = \ln^2(x) - 4\ln(x)$$ where $$\ln(x)$$ is the natural logarithm of $$x$$. 2. **Rewrite the function:** Let $$t = \ln(x)$$. Then the function becomes $$y = t^2 - 4t$$. 3. **Identify the type of function:** This is a quadratic function in terms of $$t$$. 4. **Find the vertex of the quadratic:** The vertex formula for $$y = at^2 + bt + c$$ is at $$t = -\frac{b}{2a}$$. Here, $$a=1$$ and $$b=-4$$, so $$t = -\frac{-4}{2 \times 1} = 2$$. 5. **Calculate the vertex value:** Substitute $$t=2$$ into $$y = t^2 - 4t$$: $$y = 2^2 - 4 \times 2 = 4 - 8 = -4$$. 6. **Rewrite vertex in terms of $$x$$:** Since $$t = \ln(x) = 2$$, we have $$x = e^2$$. 7. **Domain:** Since $$\ln(x)$$ is defined for $$x > 0$$, the domain is $$x > 0$$. 8. **Find intercepts:** - **y-intercept:** Not defined because $$x=0$$ is not in domain. - **x-intercepts:** Set $$y=0$$: $$t^2 - 4t = 0 \Rightarrow t(t-4) = 0 \Rightarrow t=0 \text{ or } t=4$$. Convert back to $$x$$: $$\ln(x) = 0 \Rightarrow x = e^0 = 1$$ $$\ln(x) = 4 \Rightarrow x = e^4$$. 9. **Summary:** - Domain: $$x > 0$$ - Vertex at $$(e^2, -4)$$ - x-intercepts at $$x=1$$ and $$x=e^4$$ This analysis helps understand the shape and key points of the function.