$\mathrm{Y}(\mathrm{x},\mathrm{t})=\frac{0.8}{\left[(4\mathrm{x}+5\mathrm{t}{)}^2+5\right]}$ represents a moving pulse where x and y are in meter and t is in second. Then

The shape of pulse at $x=0\text{ and }t=0$ would be as shown in fig (a)

$y(0,0)=\frac{0.8}{5}=0.16\mathrm{m}$
From the figure it is clear that ${\mathrm{y}}_{max}=0.16\mathrm{m}$
Pulse will be symmetric (Symmetry is checked about $y_{max}\text{ ) if at }t=0y\left(x\right)=y(-x)$
From the given equation,
and $\left.\begin{array}{c}y\left(x\right)=\frac{0.8}{16x^2+5}\\ y(-x)=\frac{0.8}{16x^2+5}\end{array}\right\}\text{ at }t=0$
or $y\left(x\right)=y(-x)$
Therefore, pulse is symmetric
Speed of pulse
At $\mathrm{t}=1\mathrm{s}\text{ and }\mathrm{x}=-1.25\mathrm{m}$ value of y is again 0.16 m, ie., pulse has travelled a distance of 1.25 m in 1s in negative x-direction or we can say that the speed of pulse is 1.25 m/s and it is travelling in negative x-direction. Therefore, it will travel a distance of 2.5 m in 2 s. The above statement can be better understood from fig (b)