1. **Problem statement:** Given the logarithmic function $y = -1 + \ln(x+1)$, we need to find the asymptote lines, domain, range, x-intercept, y-intercept, and another point on the graph. 2. **Asymptote lines:** The function involves $\ln(x+1)$, which is defined only for $x+1 > 0 \Rightarrow x > -1$. The vertical asymptote occurs where the argument of the logarithm is zero: $$x + 1 = 0 \Rightarrow x = -1$$ So, the vertical asymptote is the line: $$x = -1$$ There is no horizontal asymptote for logarithmic functions. 3. **Domain and range:** - Domain: Since $x+1 > 0$, domain is: $$(-1, \infty)$$ - Range: The natural logarithm function $\ln(t)$ for $t > 0$ has range $(-\infty, \infty)$. Adding $-1$ shifts the graph down by 1, so range remains: $$(-\infty, \infty)$$ 4. **Find the x-intercept:** Set $y=0$ and solve for $x$: $$0 = -1 + \ln(x+1)$$ $$\ln(x+1) = 1$$ Exponentiate both sides: $$e^{\ln(x+1)} = e^1$$ $$x+1 = e$$ $$x = e - 1$$ 5. **Find the y-intercept:** Set $x=0$ and find $y$: $$y = -1 + \ln(0+1) = -1 + \ln(1) = -1 + 0 = -1$$ So the y-intercept is at $(0, -1)$. 6. **Another point on the graph:** Choose $x=1$: $$y = -1 + \ln(1+1) = -1 + \ln(2)$$ Numerically, $\ln(2) \approx 0.693$, so: $$y \approx -1 + 0.693 = -0.307$$ Point: $(1, -0.307)$ 7. **Summary:** - Vertical asymptote: $x = -1$ - Domain: $(-1, \infty)$ - Range: $(-\infty, \infty)$ - x-intercept: $(e-1, 0)$ - y-intercept: $(0, -1)$ - Another point: $(1, -1 + \ln(2))$