`(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`