## Calculus 10th Edition

$x$-intercepts $=0,-3$ $y$-intercept $=0$
$y=\dfrac{x^{2}+3x}{(3x+1)^{2}}$ Find the $y$-intercept by setting $x$ equal to zero and solving for $y$: $x=0$ $y=\dfrac{(0)^{2}+3(0)}{[3(0)+1]^{2}}=\dfrac{0+0}{(0+1)^{2}}=\dfrac{0}{1}=0$ $y$-intercept $=0$ Find the $x$-intercept by setting $y$ equal to zero and solving for $x$: $y=0$ $0=\dfrac{x^{2}+3x}{(3x+1)^{2}}$ $(0)(3x+1)^{2}=x^{2}+3x$ $x^{2}+3x=0$ $x(x+3)=0$ $x=0$ $x+3=0$ $x=-3$ $x$-intercepts $=0,-3$