yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Reflecting functions introduction | Transformations of functions | Algebra 2 | Khan Academy


5m read
·Nov 10, 2024

So what you see here, this is a screenshot of the Desmos online graphing calculator. You can use it at desmos.com, and I encourage you to use this after this video or even while I'm doing this video.

But the goal here is to think about the reflection of functions. So let's just start with some examples. Let's say that I had a function f of x, and it is equal to the square root of x. So that's what it looks like; fairly reasonable. Now let's make another function g of x, and I'll start off by also making that the square root of x. So no surprise there; g of x was graphed right on top of f of x.

But what would happen if, instead of it just being the square root of x, what would happen if we put a negative out front right over there? What do you think is going to happen when I do that? Well, let's just try it out. When I put the negative, it looks like it flipped it over the x-axis. It looks like it reflected it over the x-axis.

Now, instead of doing it that way, what if we had another function h of x? I'll start off by making it identical to f of x, so once again it's right over there; it traces out f of x. Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign? What if we replaced x with a negative x? What do you think is going to happen there? Well, let's try it out.

If we replace it, that shifted it over the y-axis. Then pause this video and think about how would you shift it over both axes? Well, we could do a... well, I'm running out of letters; maybe I will do a... I don't know... k of x is equal to... so I'm going to put the negative outside the radical sign, and then I'm going to take the square root and I'm going to put a negative inside the radical sign. Notice it flipped it over both; it flipped it over both the x-axis and the y-axis to go over here.

Now, why does this happen? Let's just start with the g of x. So when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it. So when x is zero, we got zero. When x is one, instead of one now, you're taking the negative of it, so you're going to get negative 1. When x is 4, instead of getting positive 2, you're now going to get negative 2. When x is equal to 9, instead of getting positive 3, you now get negative 3.

So hopefully, that makes sense why putting a negative out front of an entire expression is going to flip it over... flip its graph over the x-axis. Now, what about replacing an x with a negative x? One way to think about it now is, whatever you inputted one before, that would now be a negative one that you're trying to evaluate the principal root of. We know that the principal root function is not defined for negative one, but when x is equal to negative one, our original function wasn't defined there when x is equal to negative 1.

But if you take the negative of that, well now you're taking this principal root of 1. And so that's why it is now defined. So whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x... and on the negative value of that x. And so that's why it flips it over the y-axis.

This is true with many types of functions; we don't have to do this just with a square root function. Let's try another function. Let's say we tried this for e to the x power. So there you go; we have a very classic exponential there. Now, let's say that g of x is equal to negative e to the x, and if what we expect to happen happens, this will flip it over the x-axis.

So negative e to the x power, and indeed that is what happens. And then how would we flip it over the y-axis? Well, let's do an h of x that's going to be equal to e to the, instead of putting an x there, we will put a negative x. Negative x, and there you have it; notice it flipped it over the y-axis.

Now both examples that I just did, these are very simple expressions. Let's imagine something that's a little bit more complex. Let's say that f of x... let's give it a nice higher degree polynomial. So let's say it's x to the third minus 2 x squared; that's a nice one. And actually, let's just add another term here, so plus 2 x. And I want to make it... let me get minus 2x; I want to accentuate some of those curves.

All right, so that's a pretty interesting graph. Now how would I flip it over the x-axis? Well, the way that I would do that is I could define a g of x; I could do it two ways. I could say g of x is equal to the negative of f of x, and we get that. So that's essentially just taking this entire expression and multiplying it by negative 1.

And notice it's multiplying it; it's flipping it over the x-axis. Another way we could have done it is, instead of that, we could have said the negative of x to the third minus... minus 2x squared and then minus 2x, and then we close those parentheses, and we get the same effect.

Now, what if we wanted to flip it over the y-axis? Well, then instead of putting a negative on the entire expression, what we want to do is replace our x's with a negative x. So you could do it like this; you could say that that's going to be f of negative x, and that has the effect of everywhere you saw an x before, you replace it with a negative x. And notice it did exactly what we expect; it flipped it over the y-axis.

Now, the other way we could have done that, just to make it clear, that's the same thing as negative x to the third power minus 2 times negative x squared minus 2 times negative x. And of course, we could simplify that expression, but notice it has the exact same idea.

And if we wanted to flip it over both the x and y axes, well, we've already flipped it over the y-axis. To flip it over the x-axis—oops, I just deleted it—to flip it over the... I'm having issues here—to flip it over the x-axis as well, we would have... give me a parentheses already. I would just put a negative out front.

So I put a negative out front, and there you have it; this flipped it over both the x and y axes. You can do them in either order, and you will get to this green curve.

Now, an easier way of writing that would have been just the negative of f of negative x, and you would have gotten to that same place. So go to Desmos, play around with it. It's really good to build this intuition and really understand why it's happening.

More Articles

View All
In the Studio Pt. 2 ft Zedd | One Strange Rock
They didn’t want me to create a Zedd song. They wanted me to create a piece of music that matches what this is all about. [music playing] My first thoughts when the project came to me was, finally, and excited, because I’ve made classical music in my li…
Calculating kinetic energy | Modeling energy | High school physics | Khan Academy
In this video, we’re going to talk about kinetic energy, and we’re also going to think about how to calculate it. So, you can already imagine, based on the word “kinetic,” which is referring to motion, that this is the energy that an object has by virtue …
Average Net Worth By Age (The Sad Truth)
What’s up, guys? It’s Graham here. So, the other day, I came across an article which found that 60% of Americans are currently living paycheck to paycheck. That got me thinking: what is the average net worth throughout every age, and is that realistically…
Being a CEO (What they don’t tell you)
What’s your favorite position?” he asked. She said, “CEO.” “Are you for real?” Okay, the media glamorizes these high-power CEOs without actually revealing what goes on behind the scenes. In a recent interview with Jensen Huang, the CEO of the most valu…
Simulating robots with module imports | Intro to CS - Python | Khan Academy
Let’s design a program that imports functionality from another file. When programming teams collaborate on projects, they’re often writing code across multiple files. They package their work into functions and then share them for other team members to use…
Integration using completing the square and the derivative of arctan(x) | Khan Academy
All right, let’s see if we can find the indefinite integral of ( \frac{1}{5x^2 - 30x + 65} \, dx ). Pause this video and see if you can figure it out. All right, so this is going to be an interesting one. It’ll be a little bit hairy, but we’re going to w…