# What is the equation of the normal line of #f(x)=e^x-x^3# at #x=0#?

##### 1 Answer

Dec 20, 2015

In slope intercept form:

#y = -x+1#

#### Explanation:

Given

Then

#f(0) = e^0 - 0 = 1#

So the graph of the function passes through

#f'(0) = e^0 - 0 = 1#

So the slope of the *tangent* at

If the slope of the tangent is

The equation of a line of slope

#(y - y_0) = m(x - x_0)#

In our case we have

#(y - 1) = -1(x - 0) = -x#

That is

graph{(e^x-x^3-y)(x+y-1) = 0 [-10, 10, -5, 5]}