`(dy)/(dx)=1/(1+(x/2)^2)(1/2)`

When `x = 3`, this expression is equal to: `0.153846`

So the slope of the tangent at `x = 3` is `0.153846`.

The slope of the normal at `x = 3` is given by:

`(-1)/0.153846=-6.5`

So the equation of the normal is given by:

(When `x = 3`, `y = 0.9828`)

`y − 0.9828 = -6.5(x − 3)`

That is,

`y = -6.5x + 20.483`