## College Algebra 7th Edition

$\textbf{(a)}\hspace{0.7cm}$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 $(0,0)$.
$\textbf{(a)}\hspace{0.7cm}$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 $(0,0)$