Given:
y is inversely proportional to square of x.
The equation is written as,
[tex]\begin{gathered} y\propto\frac{1}{x^2} \\ y=\frac{c}{x^2}\ldots\ldots\ldots c\text{ is constant} \end{gathered}[/tex]
Also y = 0.25 when x = 5.
[tex]\begin{gathered} y=\frac{c}{x^2} \\ 0.25=\frac{c}{5^2} \\ 25\times0.25=c \\ c=\frac{25}{4} \end{gathered}[/tex]
So, the equation of y interms of x is,
[tex]y=\frac{25}{4x^2}[/tex]
When x increases,
[tex]\begin{gathered} \lim _{x\to\infty}y=\lim _{x\to\infty}(\frac{25}{4x^2}) \\ =\frac{25}{4}\lim _{x\to\infty}(\frac{1}{x^2}) \\ =0 \end{gathered}[/tex]
Hence, the value of x increases then y decreases.
