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

Dec 20, 2015

In slope intercept form:

$y = - x + 1$

#### Explanation:

Given $f \left(x\right) = {e}^{x} - {x}^{3}$

Then $f ' \left(x\right) = {e}^{x} - 3 {x}^{2}$

$f \left(0\right) = {e}^{0} - 0 = 1$

So the graph of the function passes through $\left(0 , 1\right)$

$f ' \left(0\right) = {e}^{0} - 0 = 1$

So the slope of the tangent at $\left(0 , 1\right)$ is $1$.

If the slope of the tangent is $m$ then the slope of the normal is $- \frac{1}{m}$. So in our example the slope of the normal is $- \frac{1}{1} = - 1$.

The equation of a line of slope $m$ passing through a point $\left({x}_{0} , {y}_{0}\right)$ can be written as:

$\left(y - {y}_{0}\right) = m \left(x - {x}_{0}\right)$

In our case we have $\left({x}_{0} , {y}_{0}\right) = \left(0 , 1\right)$ and $m = - 1$, so:

$\left(y - 1\right) = - 1 \left(x - 0\right) = - x$

That is $y = - x + 1$ in slope intercept form.

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