Question

what is a line perpendicular to Fx=[(-e^-1)*x-(e^-1)]?

Answer 1

\[\Large\rm F(x)=-x e^{-1}-e^{-1}\]Am I reading that correctly? Something like this?

Answer 2

yea

Answer 3

I'll put the x back behind like you had it written.
\[\Large\rm F(x)=\color{orangered}{\left(-e^{-1}\right)}x-e^{-1}\]It will make more sense in this form.
Remember your slope-intercept form of a line?\[\Large\rm y(x)=\color{orangered}{m}x+b\]

Answer 4

The coefficient on our x is the slope, we can recognize that from the formula, yes?

Answer 5

i think so but im not sure since its part of a longer question

Answer 6

For a line to be `perpendicular` to this one,
it means the slope will be the `negative reciprocal` of this slope.

So we have some other line that we're trying to form:\[\Large\rm y=mx+b\]And the slope will be the negative, and flip, of our slope from F(x).

Answer 7

okay, i get that but i get confused when working with e

Answer 8

Yes this is going to be weird :)
We're working with `e`, and we have to pay attention to our exponent rules very carefully in this next step.

Answer 9

So we take our slope,
\[\Large\rm -e^{-1}~negative\to\qquad =e^{-1}~and~flip\to\qquad=\frac{1}{e^{-1}}\]

Answer 10

okay

Answer 11

Then we want to remember our exponent rule,
to take the negative off of the power we toss him up into the numerator, yes?\[\Large\rm \frac{1}{e^{-1}}=\frac{e^{+1}}{1}\]

Answer 12

yea

Answer 13

Ignore the denominator, since it's just one.
Ignore the exponent since it's just one...
So our new slope, our perpendicular line's slope, is simply \(\Large\rm m=e\)

Answer 14

Fancy stuff huh? :O

Answer 15

oh thanks XD

Answer 16

c: