1. **State the problem:** We are given the function $$y=\frac{1}{x-3}$$ and want to understand its behavior and key features. 2. **Formula and rules:** This is a rational function with a denominator that cannot be zero. The function is undefined at $$x=3$$ because division by zero is undefined. 3. **Find the vertical asymptote:** Set the denominator equal to zero: $$x-3=0$$ $$x=3$$ So, there is a vertical asymptote at $$x=3$$. 4. **Find the horizontal asymptote:** For large $$|x|$$, the term $$x-3$$ behaves like $$x$$, so: $$y=\frac{1}{x-3} \approx \frac{1}{x}$$ As $$x \to \pm \infty$$, $$y \to 0$$. Thus, the horizontal asymptote is $$y=0$$. 5. **Find intercepts:** - **x-intercept:** Set $$y=0$$: $$0=\frac{1}{x-3}$$ This equation has no solution because $$\frac{1}{x-3}$$ is never zero. So, no x-intercept. - **y-intercept:** Set $$x=0$$: $$y=\frac{1}{0-3}=\frac{1}{-3}=-\frac{1}{3}$$ So, the y-intercept is at $$\left(0,-\frac{1}{3}\right)$$. 6. **Summary:** - Vertical asymptote at $$x=3$$ - Horizontal asymptote at $$y=0$$ - No x-intercept - y-intercept at $$\left(0,-\frac{1}{3}\right)$$ This explains the behavior of the graph near the asymptotes and intercepts. **Final answer:** The function $$y=\frac{1}{x-3}$$ has a vertical asymptote at $$x=3$$, a horizontal asymptote at $$y=0$$, no x-intercept, and a y-intercept at $$\left(0,-\frac{1}{3}\right)$$.