1. **State the problem:** We want to analyze the function $$y=4+\frac{1}{\sqrt{2-x}}$$ and understand its behavior. 2. **Identify the domain:** The expression under the square root must be positive for the function to be real and defined. So, $$2-x>0 \implies x<2$$. 3. **Rewrite the function:** The function is $$y=4+\frac{1}{\sqrt{2-x}}$$. 4. **Behavior near the domain boundary:** As $$x \to 2^-$$, $$\sqrt{2-x} \to 0^+$$, so $$\frac{1}{\sqrt{2-x}} \to +\infty$$, thus $$y \to +\infty$$. 5. **Behavior as $$x \to -\infty$$:** $$\sqrt{2-x} \approx \sqrt{-x}$$ grows large, so $$\frac{1}{\sqrt{2-x}} \to 0^+$$, thus $$y \to 4^+$$. 6. **No intercepts:** Since $$\sqrt{2-x}$$ is always positive in the domain, $$\frac{1}{\sqrt{2-x}}$$ is positive, so $$y>4$$ always. The function never crosses the x-axis or y-axis at zero. 7. **Summary:** The function is defined for $$x<2$$, increases without bound as $$x \to 2^-$$, and approaches 4 from above as $$x \to -\infty$$. Final answer: The function $$y=4+\frac{1}{\sqrt{2-x}}$$ is defined for $$x<2$$ and has a vertical asymptote at $$x=2$$ with $$y \to +\infty$$, and a horizontal asymptote at $$y=4$$ as $$x \to -\infty$$.