1. **State the problem:** We need to sketch the functions $$f(x) = \frac{x^2 - 9}{x + 3}$$ and $$g(x) = \frac{2x}{x - 1}$$ on the same coordinate grid with $$x$$ from about -5 to 5 and $$y$$ from about -30 to 30. We must label asymptotes and intercepts and shade the regions on the $$x$$-axis where $$f(x) < g(x)$$. 2. **Simplify and analyze $$f(x)$$:** Notice that $$x^2 - 9 = (x - 3)(x + 3)$$, so $$f(x) = \frac{(x - 3)(x + 3)}{x + 3}$$ For $$x \neq -3$$, we can cancel $$x + 3$$: $$f(x) = \cancel{\frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}}} = x - 3$$ However, $$x = -3$$ is a vertical asymptote (removable discontinuity) because the original denominator is zero there. 3. **Analyze $$g(x)$$:** The function is $$g(x) = \frac{2x}{x - 1}$$ Vertical asymptote at $$x = 1$$ (denominator zero). 4. **Find intercepts:** - For $$f(x)$$: - **x-intercept:** Set $$f(x) = 0$$, so $$x - 3 = 0 \Rightarrow x = 3$$. - **y-intercept:** Evaluate $$f(0) = 0 - 3 = -3$$. - For $$g(x)$$: - **x-intercept:** Set numerator zero: $$2x = 0 \Rightarrow x = 0$$. - **y-intercept:** Evaluate $$g(0) = \frac{0}{-1} = 0$$. 5. **Asymptotes:** - $$f(x)$$ has a vertical asymptote at $$x = -3$$. - $$g(x)$$ has a vertical asymptote at $$x = 1$$. 6. **Compare $$f(x)$$ and $$g(x)$$ to find where $$f(x) < g(x)$$:** Set inequality: $$x - 3 < \frac{2x}{x - 1}$$ Multiply both sides by $$x - 1$$ (consider sign): $$ (x - 3)(x - 1) < 2x $$ Expand left side: $$x^2 - x - 3x + 3 < 2x$$ Simplify: $$x^2 - 4x + 3 < 2x$$ Bring all terms to one side: $$x^2 - 4x + 3 - 2x < 0$$ $$x^2 - 6x + 3 < 0$$ 7. **Solve quadratic inequality:** Find roots of $$x^2 - 6x + 3 = 0$$ using quadratic formula: $$x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2\sqrt{6}}{2} = 3 \pm \sqrt{6}$$ Approximate roots: $$3 - \sqrt{6} \approx 0.55, \quad 3 + \sqrt{6} \approx 5.45$$ Since the parabola opens upward, inequality $$x^2 - 6x + 3 < 0$$ holds between the roots: $$0.55 < x < 5.45$$ 8. **Consider domain restrictions:** - $$f(x)$$ undefined at $$x = -3$$. - $$g(x)$$ undefined at $$x = 1$$. Check intervals avoiding vertical asymptotes. 9. **Final answer:** - Sketch $$f(x) = x - 3$$ with a hole at $$x = -3$$. - Sketch $$g(x) = \frac{2x}{x - 1}$$ with vertical asymptote at $$x = 1$$. - Label intercepts: $$f$$ at (3,0) and (0,-3), $$g$$ at (0,0). - Label vertical asymptotes at $$x = -3$$ and $$x = 1$$. - Shade the region on the $$x$$-axis where $$f(x) < g(x)$$, which is approximately $$0.55 < x < 5.45$$.