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

##### 1 Answer

Jan 31, 2017

#### 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

so If

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

When

# \ \ \ \y=e^0=1 #

# dy/dx=-e^0=-1 #

So the tangent passes through

# 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]}