Question

Exercise 2 Find a formula for Y in terms of X

Answer 1

Given:

y is inversely proportional to square of x.

The equation is written as,

`\begin{gathered} y\propto(1)/(x^2) \\ y=(c)/(x^2)\ldots\ldots\ldots c\text{ is constant} \end{gathered}`

Also y = 0.25 when x = 5.

`\begin{gathered} y=(c)/(x^2) \\ 0.25=(c)/(5^2) \\ 25*0.25=c \\ c=(25)/(4) \end{gathered}`

So, the equation of y interms of x is,

`y=(25)/(4x^2)`

When x increases,

`\begin{gathered} \lim _(x\to\infty)y=\lim _(x\to\infty)((25)/(4x^2)) \\ =(25)/(4)\lim _(x\to\infty)((1)/(x^2)) \\ =0 \end{gathered}`

Hence, the value of x increases then y decreases.