Answer
$\textbf{(a)}\hspace{0.7cm}$See the graph
$\textbf{(b)}\hspace{0.7cm}$From the graph:
the $x−$intercept is at $(0,0)$ , and the y−intercept is at $(0,0)$.
$\textbf{(c)}\hspace{0.7cm} x−$intercept is at $(0,0)$ and $y−$intercept is $(0,0)$.
Work Step by Step
$\textbf{(a)}\hspace{0.7cm}$See the graph
$\textbf{(b)}\hspace{0.7cm}$From the graph:
the $x−$intercept is at $(0,0)$, and the $y−$intercept is at $(0,0)$.
$\textbf{(c)}\hspace{0.7cm}$
at $x−$intercepts $y=0$
i.e. $\dfrac{x}{x^2+1}=0⇒x=0$
i.e $x−$intercept is at $(0,0)$
at $y−$intercepts $x=0$
i.e. $y=\dfrac{0}{0^2+1}=\dfrac{0}{1}=0$
i.e$ y−$intercept is $(0,0)$
