1. **Problem Statement:** We need to analyze the function $$f(x) = -2\log_3(x + 4)$$ and find points to sketch the graph, label asymptotes, intercepts, and transformations. 2. **Basic Function:** The basic logarithmic function is $$y = \log_3 x$$. 3. **Transformations:** - Horizontal shift left by 4 units: $$x \to x + 4$$. - Vertical reflection about the x-axis: multiply by $$-1$$. - Vertical stretch by a factor of 2: multiply by $$2$$. So, $$f(x) = -2\log_3(x + 4)$$ is the transformed function. 4. **Domain:** The argument of the log must be positive: $$x + 4 > 0 \implies x > -4$$. 5. **Range:** Logarithmic functions have range $$(-\infty, \infty)$$, and vertical stretch/reflection does not change this, so range is $$(-\infty, \infty)$$. 6. **Vertical Asymptote:** At $$x = -4$$ because the log argument approaches zero. 7. **Intercepts:** - **x-intercept:** Set $$f(x) = 0$$: $$0 = -2\log_3(x + 4) \implies \log_3(x + 4) = 0 \implies x + 4 = 3^0 = 1 \implies x = -3$$. - **y-intercept:** Set $$x = 0$$: $$f(0) = -2\log_3(0 + 4) = -2\log_3 4$$. Calculate $$\log_3 4 = \frac{\ln 4}{\ln 3} \approx 1.2619$$, so $$f(0) \approx -2 \times 1.2619 = -2.5238$$. 8. **Additional Points:** - At $$x = -2$$: $$f(-2) = -2\log_3(2) = -2 \times \frac{\ln 2}{\ln 3} \approx -2 \times 0.6309 = -1.2618$$. - At $$x = -1$$: $$f(-1) = -2\log_3(3) = -2 \times 1 = -2$$. 9. **Summary of points to plot:** - Vertical asymptote at $$x = -4$$. - Intercept at $$(-3, 0)$$. - Points: $$(-2, -1.26)$$, $$(-1, -2)$$, $$(0, -2.52)$$. 10. **Labeling:** - Label vertical asymptote $$x = -4$$. - Label intercept $$(-3, 0)$$. - Label points $$(-2, -1.26)$$, $$(-1, -2)$$, and $$(0, -2.52)$$. These points and labels will help sketch the graph accurately.