In Exercises 5–9, graph f and g in the same rectangular coordinat...

Question

In Exercises 5–9, graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each function's domain and range. f(x) = 3^x and g(x) = -3^x

Accepted Answer

Welcome back. I am so glad you're here. We are given two functions. The first F of X equals two raised to the X and the second is G of X equals negative two raised to the X. And we are to apply tran transformations after we have graphed F of X in order to graph G of X. Then were to determine the assam. Totes of both F of X and G of X. And then we get to determine the domain and a range of both functions. So we've got quite a bit to do here. The first thing we're going to do is graph our F of X function. We've got a horizontal X and a vertical Y axis. And then our F. Of X is equal to two raised to the X. You might just know exactly what to raise to the X looks like wonderful. But if you don't, that's totally fine to what you can do in that situation is you can plot a few points, you can see what's going on. I'm going to plot where X equals and negative one. So plugging in one for our X to raise to the first. Well two raised to the first R. F X. There is going to equal to. So our Y value is to now graphing that we look starting at our origin where the X. And the Y axis meat. And then we're going to move one to the right for our X value of one. And then we're going to go up to for our Y value of two. So that's going to be approximately here on our graph. Now, plotting for X equals zero. That'll be two raised to the zero. Well, two raised to the zero. Anything raised to the zero equals one. So where X equals one. Y. Excuse me, where X equals zero, Y equals one. Looking at our graph starting at the origin, going staying at zero for X value and going up one for our Y value puts us approximately here. Now plugging in a negative one for X. That will be two raised to the negative one. Well, because we have that negative in our exponents, that two goes down to the denominator. So it will be 1/2 to the first. Well two to the first is to so this will be one half Graphing this starting at the origin, going to the left, one for our x value for our negative one and then we'll go up one half for our Y value of one half. So we've got these three points plotted what's happening before and after these three points. Well, we notice that as X becomes negative, we're going to have fractions. And so if we had something like X being negative 1000, if two raised to the negative 1000, that would be one over two to the 1000. That's a really big number in the denominator. But because it's a fraction, it's a really small number. So because we'll just have fractions getting closer and closer to zero that always approach zero but they'll never quite reach it. So that's what's happening as we move to the left. How about as X gets bigger? Well, let's say we had a very big number for our X value to raise to the let's say 1000 to raise to the 1000 is a huge number. So because we have exponents, this is going up into the right and it's going up pretty quickly. So that's roughly what's going on for our F of X. Now we need to apply transformations for our G of X. G of X equals negative two raised to the X. What's different between our F. Of X and our G of X. Well this negative in front of our two because the negative is not being applied directly to the X. That negative indicates that there is reflection on the X axis. If you know what that means. Great. Wonderful. If you're not sure exactly which one would indicate reflection on the X axis, which way does that go? You could plot a point To see where this function is hanging out and we could plot where x equals one. And let's plug that in. So we've got negative two raised to the first while Two to the first because we do the exponents first to to the first is to and so this negative stays, it's negative two. So where X equals one, Y equals negative two. Let's graph that starting at the origin, we're going to move one to the right for our X value of X equals one. And then we'll move down to for our Y value of X equals negative two. So when we say that there's reflection on the X axis, that means that this function is going to occur on the other side of the X axis. It will probably look a little bit better than that. But that's the general direction. All right. So we've got our F of X and G of X plotted. Now, what are the ASM totes for both of these? Well, we already talked about how when we have negative values for X, they're going to be fractions for our Y values and they're going to get so small but they will never quite reach zero, meaning that we have an awesome tote where why equals zero. So there's our ASM tote. And now for the domain and range of both functions. Well, the domain is going to be all the possible X values. And looking from left to right. X could be anything. It could be all real numbers for both of our functions. But for our ranges we have uh different ranges for F of X and G of X for F of X, it's going to have to be greater than zero. R. Y values are going to be greater than zero because they can't hit that ASM tote at zero for G of X. All of our Y values are going to be less than zero. They're all existing below the X axis. So there is our graph and our ASM totes and our domain and range. So now what do we do? Well, let's pick our um Y intercept for F of X. Where we're looking at are F of X line and our Y intercept is at 01. So let's see which graphs include that. Um Now looking at graph A we do have a an intercept at 01. Wonderful that one could qualify. But now let's look at what's going on with our G of X. R. G of X needs to be below the X axis and nope, that's not what's happening here for A. And so that answer choice does not work. All right, let's go down and see what's happening with answer choice. B. We need to have an X. F. Of X. Y intercept at 01. And here, nope. Our F. Of X is way up here at 03. So B does not work. Let's check for C. We're looking for an f. of x intercept at 01 and yes, this one does. And is our G of X reflected on the other side of the X axis? Yes, it is. So this one has potential. Let's check real quick with our answer choice. D. Alright, this one does it have an X intercept or a Y intercept for our F of X at 01? No, it doesn't. So this one doesn't work either. So see works. Let's see if it has all of the other information. The graph looks right. The ASM toad is at Y equals zero. We have the domain of F all real numbers, domain of G all real numbers. That's correct. The range of F Y is greater than zero and the range of G Y is less than zero. This works. So answer choice C. It is well done. We'll catch you on the next one.

In Exercises 5–9, graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each function's domain and range. f(x) = 3^x and g(x) = -3^x

Key Concepts

Exponential Functions

Transformations of Functions

Asymptotes

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