### Home > CALC > Chapter 1 > Lesson 1.4.1 > Problem1-141

1-141.

$\text{Visualize }f(x).\text{ It is a horizontal shift of }y=\frac{1}{x},$

which has a vertical asymptote at x = 0. f(x) also has a vertical asymptote. Where is it, and how will the vertical asymptote restrict the domain?

$\text{Find the domain of }\left ( \frac{1}{\sqrt{x-4}+2} \right )$

Both the square root and the denominator have restricted domains. Combine.

$\text{Visualize }g(x).\text{ It is a horizontal shift of }y=\sqrt{x},$

which does not exist for negative values of x. What can you conclude about g(x).

Refer to hints in part (c).