1. **Problem Statement:** 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. 2. **Identify key features:** - For $$f(x)$$, factor numerator: $$x^2 - 9 = (x - 3)(x + 3)$$. - Simplify $$f(x)$$ where possible: $$f(x) = \frac{(x - 3)(x + 3)}{x + 3}$$ - Cancel common factor $$x + 3$$ (except at $$x = -3$$ where function is undefined): $$f(x) = \cancel{\frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}}} = x - 3, \quad x \neq -3$$ 3. **Asymptotes and discontinuities:** - For $$f(x)$$: - Vertical asymptote at $$x = -3$$ (removable discontinuity, hole). - Since simplified to $$x - 3$$ except at $$x = -3$$, no vertical asymptote but a hole there. - No horizontal asymptote (degree numerator > denominator). - For $$g(x)$$: - Vertical asymptote at $$x = 1$$ (denominator zero). - Horizontal asymptote at $$y = 0$$ (degree numerator = degree denominator, leading coefficients ratio is 2/1 = 2, so actually horizontal asymptote is $$y=2$$). 4. **Intercepts:** - For $$f(x)$$: - x-intercept: set $$f(x) = 0$$, so $$x - 3 = 0 \Rightarrow x = 3$$. - y-intercept: set $$x=0$$, $$f(0) = 0 - 3 = -3$$. - For $$g(x)$$: - x-intercept: set numerator zero $$2x = 0 \Rightarrow x=0$$. - y-intercept: set $$x=0$$, $$g(0) = 0$$. 5. **Inequality $$f(x) < g(x)$$:** - Use simplified $$f(x) = x - 3$$ for $$x \neq -3$$. - Solve $$x - 3 < \frac{2x}{x - 1}$$. 6. **Solve inequality:** Multiply both sides by $$x - 1$$ (consider sign): $$ (x - 3)(x - 1) < 2x $$ Expand left: $$ x^2 - x - 3x + 3 < 2x $$ $$ x^2 - 4x + 3 < 2x $$ Bring all terms to one side: $$ x^2 - 4x + 3 - 2x < 0 $$ $$ x^2 - 6x + 3 < 0 $$ 7. **Find roots of quadratic:** $$ x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2\sqrt{6}}{2} = 3 \pm \sqrt{6} $$ 8. **Determine intervals:** - Quadratic opens upward (coefficient of $$x^2$$ positive). - Inequality $$< 0$$ holds between roots: $$ 3 - \sqrt{6} < x < 3 + \sqrt{6} $$ 9. **Consider domain restrictions:** - $$x \neq -3$$ for $$f(x)$$. - $$x \neq 1$$ for $$g(x)$$. 10. **Summary:** - Plot $$f(x) = x - 3$$ with a hole at $$x = -3$$. - Plot $$g(x) = \frac{2x}{x - 1}$$ with vertical asymptote at $$x=1$$ and horizontal asymptote at $$y=2$$. - Shade region on x-axis where $$f(x) < g(x)$$, i.e., $$x \in (3 - \sqrt{6}, 1) \cup (1, 3 + \sqrt{6})$$ (excluding $$x=1$$). **Final answer:** - Functions sketched with asymptotes and intercepts labeled. - Shaded region on x-axis between $$3 - \sqrt{6}$$ and $$3 + \sqrt{6}$$ excluding $$x=1$$.