1. **Problem Statement:** We are given the hyperbolic function $$f(x)=\frac{1}{x-2}+3$$ and need to find its domain, range, intercepts, asymptotes, axes of symmetry, and sketch the graph. 2. **Domain:** The function is undefined where the denominator is zero, so solve $$x-2=0$$ which gives $$x=2$$. Thus, the domain is all real numbers except $$x=2$$. 3. **Range:** The function is a transformation of $$y=\frac{1}{x}$$ shifted right by 2 and up by 3. The horizontal asymptote is $$y=3$$, so the range is all real numbers except $$y=3$$. 4. **Intercepts:** - **x-intercept:** Set $$f(x)=0$$: $$0=\frac{1}{x-2}+3$$ $$\frac{1}{x-2}=-3$$ $$x-2=-\frac{1}{3}$$ $$x=2-\frac{1}{3}=\frac{5}{3}$$ So, x-intercept is $$\left(\frac{5}{3},0\right)$$. - **t-intercept (y-intercept):** Set $$x=0$$: $$f(0)=\frac{1}{0-2}+3=\frac{1}{-2}+3=-\frac{1}{2}+3=\frac{5}{2}$$ So, y-intercept is $$\left(0,\frac{5}{2}\right)$$. 5. **Asymptotes:** - **Vertical asymptote:** At $$x=2$$ because the function is undefined there. - **Horizontal asymptote:** At $$y=3$$ because as $$x\to \pm \infty$$, $$f(x)\to 3$$. 6. **Axes of Symmetry:** The function is a shifted hyperbola. The original $$y=\frac{1}{x}$$ is symmetric about the origin. After shifting right by 2 and up by 3, the center of symmetry is at $$\left(2,3\right)$$. So the axes of symmetry are the lines through this point, but the function itself does not have symmetry about the x- or y-axis. 7. **Graph Sketch:** The graph has two branches separated by the vertical asymptote $$x=2$$. The left branch approaches $$x=2$$ from the left and $$y=3$$ from below. The right branch approaches $$x=2$$ from the right and $$y=3$$ from above. The intercepts and asymptotes guide the shape. **Final answers:** - Domain: $$\{x \in \mathbb{R} : x \neq 2\}$$ - Range: $$\{y \in \mathbb{R} : y \neq 3\}$$ - x-intercept: $$\left(\frac{5}{3},0\right)$$ - y-intercept: $$\left(0,\frac{5}{2}\right)$$ - Vertical asymptote: $$x=2$$ - Horizontal asymptote: $$y=3$$ - Center of symmetry: $$\left(2,3\right)$$