College Algebra: Reflecting Functions

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Okay, so let's take a look at a very specific example, or a couple of specific examples of how to find the related functions that have been reflected over various axes. So let's start off with the following: How about f(x) = 3x + 1? What does the graph of that look like? Well, that's just a straight line. In fact, it's in slope-intercept form. I could sketch a little picture of it right here for you. So, its y intercept is 1, and the slope is 3/1, so 1 over, 3 up. So there's the graph of that line.
Let's see if we can first of all find the equation of the line that is reflected over the x-axis. So what would I do? Let's just think through this. I want to reflect over the x-axis. That means that the y that used to be positive, I now want them to be negative, and the y's that used to be negative, I now want them to be positive. So I should consider here a new function, and I'll call it maybe a g, and it just equals negative the other function. And what would that equal? It would equal -3x + 1. So that should be the function, right? Is that what you think? Actually, that is absolutely wrong. And in fact, this is a classic mistake. In fact, this is classic mistake number 4 out of my top ten list. This is number 4. That's right, it's the subtracting mistake. If you're subtracting a whole bunch of stuff you have to remember to share the negativity. All I'm doing here is taking -3x and I'm still adding the 1. I've got to subtract that 1. So, in fact, a classic mistake. You must put those parentheses. Please make that mistake with me now and then never again. So, in fact, the actual answer is -3x - 1.
Let's see what the graph of that looks like and see if g really is the reflection over the x-axis. Well, this is still a line; that's a good sign, and it's intercept is now -1. Oh, that's happy, because here it was +1, now it's -1. Oh, I like that. And now the slope is -3/1. That means that I go one over and minus three, so I go down three. Do you see how that's the exact opposite point here? So, in fact, this really is the reflection of this over that. Neat! And that's how you actually find--there's the actual equation. What if someone asked you, like me, to find the equation of this line if it's reflected over the y-axis? What do you do then? Well, then what I want to do is what? I want to take all the negative values of x and flip them with the positive values of x. So in that case, let me call that h, a different function. What I do there is take f(-x). It's a different thing than this. Do you see where the negative signs are? It's a little bit different. And what would this look like? Here, wherever I see an x I replace it by -x. So I don't do anything with that 1, but here I put in 3(-x + 1). And if you work that out--there's no distributing here--I see this equals -3x + 1.
Is that really the flip over this axis? Let's see. The intercept is 1. Oh, that's good news, because nothing should change there, and the slope is now -3/1. So what does that mean? That means that I go 1 over and then -3 down. If you notice--look at that--that is exactly the split. If you take a look at this picture and look at that line and flip it over this axis, you'd see exactly this. This goes to this; this goes to that. Whereas this green, is the reflection over the other axis. So this sort of indicates and illustrates how you would actually do this reflection. If you want to reflect over the x-axis, you look at -f(x) and compute that function. If you want to reflect over the y-axis, you consider f(-x) and carefully compute that, and notice that they will not be the same, necessarily. They're different kinds of functions.
Okay, anyway, you can try your hand at some of these and see that you can actually write down the function that corresponds to the reflection over the x or the y-axis. Neat! Aren't we having fun? I am.
Relations and Functions
Manipulating Graphs- Symmetry and Reflections
Reflecting Specific Functions Page [1 of 1]