The function given is,
[tex]f(x)=\frac{1}{\sqrt[]{x+1}}[/tex]
The graph of this function:
We can also find the asymptotes the following way:
Horizontal Asymptote
[tex]\begin{gathered} f(x)=\frac{1}{\sqrt[]{x+1}} \\ y=\frac{1}{\sqrt[]{x+1}} \\ y=\frac{\frac{1}{x}}{\sqrt[]{\frac{x}{x}+\frac{1}{x}}} \\ y=\frac{\frac{1}{x}}{\sqrt[]{1+\frac{1}{x}}} \\ as \\ x\rightarrow\infty \\ y\rightarrow0 \end{gathered}[/tex]
As for Vertical Asymptote,
[tex]\begin{gathered} x+1\neq0 \\ x\neq-1 \\ ------- \\ VA\rightarrow x=-1 \end{gathered}[/tex]
(a)
From the graph, we can see that horizontal asymptote is at y = 0
And, the vertical asymptote is at x = -1
The graph below (with asymptotes showing):
(b)
As seen from the graph, the function is continuous from x > -1
There aren't any function at x = -1 and less
So, the continuity interval is
[tex]x>-1[/tex]
(c)
The function is discontinuous at x = -1
This is an infinite discontinuity because it's a vertical asymptote.
