Explaining the Reflection of e^x at y=2

[brycenrg]

Reflect e^x at y=2


If you reflect e^x at y=0. You just turn e^x negative and it reflects



On my homework it says the correct answer is 4 - e^x
But it makes more sense for me to say its 2 - e^x.
Can someone explain?

 

[SammyS]

Reflect e^x at y=2

If you reflect e^x at y=0. You just turn e^x negative and it reflects

On my homework it says the correct answer is 4 - e^x
But it makes more sense for me to say its 2 - e^x.
Can someone explain?

Graph those functions.

 

[brycenrg]

I did, 4 - e^x is reflected at y = 4

 

[SammyS]

I did, 4 - e^x is reflected at y = 4

Does this mean you now understand what's going on ?

 

[brycenrg]

No :(. So if I want to reflect e^x at y = 5 i still do the same thing and add 5 and give it a reflection -e^x.
I don't see it

 

[SammyS]

No :(. So if I want to reflect e^x at y = 5 i still do the same thing and add 5 and give it a reflection -e^x.
I don't see it

Let's go back to the original problem.

Reflect e^x at y=2 .​

First of all, when I asked about graphing the function, I was referring to all three functions.

y = e^x

y = 2 - e^x

y = 4 - e^x​

The following procedure may help you understand the correct answer.

First shift y = e^x down 2 units.

Then take the negative (take the opposite) of that.

Then shift the result up by 2 units.(This thread likely should be moved to pre-Calculus.)

 

[Mark44]

No :(. So if I want to reflect e^x at y = 5 i still do the same thing and add 5 and give it a reflection -e^x.
I don't see it

It might help your understanding to think more in line with what's actually happening. You're reflecting the curve y = e across the horizontal line y = 5. When you say you want to reflect e^x at y = 5, I'm not sure you're understanding how this reflection is working.

 

[Mentallic]

We have a thread on this exact same problem:

https://www.physicsforums.com/showthread.php?t=734702&highlight=reflection

It would be best if you drew these functions y=e^x, y=2, y=2-e^x, y=4-e^x for yourself because then it would become painfully clear what's going on.

 

[ehild]

@brycenrg: You need to understand the concept of reflection. See http://www.mathsisfun.com/geometry/reflection.html

Looking into a mirror, you see your image behind the mirror at the same distance as the distance between you and the mirror.

A point P and its mirror image P' are at equal distances from the mirror line, at opposite sides. See attachment. If g(x) (green curve) is the mirror image of the function f(x) (blue curve) with respect to the line y=2, the point P (x, f(x)) and P' (x, g(x)) are at equal distances from y=2, one below, the other above the line: 2-f(x)= g(x)-2.

ehild