Answer
$y=\pm x \sqrt {x+1}$
Also, see the attached graph for $y=\pm x \sqrt {x+1}$.
Work Step by Step
Calculate $f(t)$.
This shows that it is a vertical parabola with zeroes at $t=\pm 1$
$f(t)=k(t-1)(t+1)=k(t^2-1)$
Refer graph, we get $f(0)=-1$ hence $k=1$
$$f(t)=(t^2-1)=x ... (1)$$
we can also write $t=\pm \sqrt {x+1}$ ... (2)
Now we will calculate $g(t)$ which shows that it is a cubical polynomial with zeroes at $t=0,\pm 1$
$g(t)=k(t-1)(t+1)=k(t^2-1)$
$g(0.5)=-0.4$ ( from graph)
Thus, $$k=1$$
$g(t)=(t^2-1)=y$
It can be written as:$y=\pm x \sqrt {x+1}$
Also, see the attached graph for $y=\pm x \sqrt {x+1}$.
