Question

How do we find the range of the binary relation `4x^2 + 9y^2 = 36` ?

Answer 1

Set x equal to zero and then solve the equation for both values of y; the negative value will be the minimum of the range and the positive value will be the maximum.

Explanation:

Given: `4x^2 + 9y^2 = 36`

Let `x = 0`:

`4(0)^2 + 9y^2 = 36`

`9y^2=36`

`y^2=4`

`y = -2 and y = 2`

The range is `-2 <= y <= 2`

Answer 2

The range is `y in [-2,2]`

Explanation:

The equation is

`4x^2+9y^2=36`

`4x^2=36-9y^2`

`x^2=9/4(4-y^2)`

`x=3/2sqrt(4-y^2)`

`x=3/2sqrt((2-y)(2+y))`

`x=f(y)`

`x` will depend on `(2-y)(2+y)>=0`

Let `p(y)=(2-y)(2+y)`

We need a sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-2` `color(white)(aaaaaaaa)` `2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `2+y` `color(white)(aaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `2-y` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `-`

`color(white)(aaaa)` `p(y)` `color(white)(aaaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `-`

Therefore,

`p(y)>=0` when `y in [-2,2]`

graph{4x^2+9y^2=36 [-9.22, 8.56, -3.235, 5.654]}