Answer
Graph 2
Work Step by Step
When $t=0$ then
$x=t+\sin 2t=0;y=t+\sin 3t=0$ so the graph passes from origin (0,0)
And when
$t\rightarrow \infty \Rightarrow x=\lim _{t\rightarrow \infty }\left( t+\sin 2t\right) =t\left( 1+\dfrac {\sin 2t}{t}\right) =\infty $
$t\rightarrow \infty \Rightarrow y=\lim _{t\rightarrow \infty }\left( t+\sin 3t\right) =t\left( 1+\dfrac {\sin 3t}{t}\right) =\infty $
$t\rightarrow -\infty \Rightarrow x=\lim _{t\rightarrow -\infty }\left( t+\sin 2t\right) =t\left( 1+\dfrac {\sin 2t}{t}\right) =-\infty $
$t\rightarrow -\infty \Rightarrow y=\lim _{t\rightarrow -\infty }\left( t+\sin 3t\right) =t\left( 1+\dfrac {\sin 3t}{t}\right) =-\infty $
So this is graph 2 (when $x\rightarrow \infty \Rightarrow y\rightarrow \infty ;x\rightarrow -\infty \Rightarrow y\rightarrow -\infty $ )
