Question

How do you find the equations for the normal line to `y=e^-x` through (0,0)?

Answer 1

`y = x+1`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is `-1`

so If `y = e^(-x)` then differentiating wrt gives us:

`dy/dx = -e^(-x)`

When `x = 0 =>`

`\ \ \ \y=e^0=1`
`dy/dx=-e^0=-1`

So the tangent passes through `(0,1)` and has gradient `-1`, hence the normal has gradient `-1` so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek is;

`y-1 = 1(x-0)`
`:. y-1 = x`
`:. y = x+1`

We can confirm this solution is correct graphically:
graph{(y-e^(-x))(y-x-1)=0 [-15, 15, -10, 10]}