The displacement equation of a travelling wave pulse, moving along ‘x’ is given by $\mathrm{Y}=\frac{8}{(4+{\mathrm{X}}^2)}$ at $t=0$ and $\mathrm{y}=\frac{8}{({\mathrm{x}}^2-10\mathrm{x}+29)}$  at $t=1$, the speed of the pulse is

The given equation corresponding to $\mathrm{t}=1$ can be rewritten as

 $\mathrm{Y}=\frac{8}{({\mathrm{X}}^2-10\mathrm{X}+25)+4}\Rightarrow \mathrm{Y}=\frac{8}{4+{(\mathrm{X}-5)}^2}$,

 $\therefore \mathrm{Y}=\frac{8}{4+{(\mathrm{X}-5\left(1\right))}^2}$

Since the displacement equation of a wave is necessarily a function of $(\mathrm{X}\pm \mathrm{Vt})$.

 So, the above equation can be rewritten as $Y=\frac{8}{[4+{(X-5t)}^2]}$ 

Evidently factor $(X-5t)$ reveals that the wave is moving along positive x-axis and its speed is 5 units