# Thread: Finding the "normal line"

1. ## Finding the "normal line"

I understand in theory how to do this...it's just the particular problem that's throwing me.

Question: Find the equation of the normal line to the graph of $f(x) = e^x - 1$ at the point where it crosses the x-axis.

My Attempt:

So $f\prime(x) = e^x$

f(x) crosses the x-axis at pretty much x = 0 I think.

so do I plug in 0 for x and solve in the original equation like this?
$
f(x) = e^0 - 1
= 0$

Giving the point (0,0)

then plug that into y = mx + b

I get

$0 = e^x(0) + b$
so $b = 0$

Does that mean the tangent line is just $y= e^x(x)$?

Making the normal line $y = -\frac{1}{e^x}(x)$ ?

Is that right? Or am I doing this completely wrong?

2. Originally Posted by lysserloo
I understand in theory how to do this...it's just the particular problem that's throwing me.

Question: Find the equation of the normal line to the graph of $f(x) = e^x - 1$ at the point where it crosses the x-axis.

My Attempt:

So $f\prime(x) = e^x$

f(x) crosses the x-axis at pretty much x = 0 I think.

so do I plug in 0 for x and solve in the original equation like this?
$
f(x) = e^0 - 1
= 0$

Giving the point (0,0)

then plug that into y = mx + b

I get

$0 = e^x(0) + b$
so $b = 0$

Does that mean the tangent line is just $y= e^x(x)$?

Making the normal line $y = -\frac{1}{e^x}(x)$ ?

Is that right? Or am I doing this completely wrong?
Note that $f^{\prime}(0)=e^0=1$.

So $y=x$ is the tangent line at that point, which then implies $y=-x$ is the normal line at that point.