Answer
The graph for $y=\pm x \sqrt {x+1}$ is as shown.
As $t$ increases, the curve is traced from the bottom right, to the top left, then bottom left, and finally top right.
Work Step by Step
Finding $f(t)$:
It is a vertical parabola with zeroes at $t=\pm 1$
thus, $f(t)=k(t-1)(t+1)=k(t^2-1)$
from the graph, we have $f(0)=-1$ hence $k=1$
$f(t)=(t^2-1)=x$ ... (1)
we can also write $t=\pm \sqrt {x+1}$ ... (2)
Finding $g(t)$
It is a cubical polynomial with zeroes at $t=0,\pm 1$
thus, $g(t)=k(t-1)(t+1)=k(t^2-1)$
from the graph, we have $g(0.5)=-0.4$
hence $k=1$
$g(t)=(t^2-1)=y$
we can also write $y=\pm x \sqrt {x+1}$
The graph for $y=\pm x \sqrt {x+1}$ is as shown.
