In Exercises 25-34, begin by graphing f(x) = 2^x. Then use transf...

Question

In Exercises 25-34, begin by graphing f(x) = 2^x. Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn graphs. g(x) = −2^x

Accepted Answer

Hello everyone. In this video, we're going to be looking at this practice problem where we want to grab a five X which is equal to nine to the X to start. And then graph G of X which is equal to negative nine to the X after using transformations. And once we graph those figure out the ascent oats and provide the equations for those ascent oats and figure out the domain and range for the functions. So in this case to get started with graphing F of X which is equal to nine X, we can do a chart where we have the inputs and the outputs. So in this case, if I plug in I'll start with zero F of X is equal to 9 to 0, which is equal to one. So when X is equal to zero, F of X is equal to one. I plug in one, I get nine to the first power which is equal to nine. I plug in two, I get nine squared which is equal to nine times nine which is 81. Now also do negative values. So if I plug in negative one, I get nine to the negative one Which I can convert into a fraction where nine to the negative one is equal to one divided by nine to the first power. If I plug in negative two, I get nine to the negative two, which is equal to one divided by nine Squared. And I can do that multiplication. So one divided by nine squared is the same as 1/81. So now I have a couple values for F of X. So I can draw a graph and or drawl the Y access and then the X axis And I'll say this is negative one negative two, Positive one positive too. And for my Y axis, I'll go Up in nine. So 9 And 18 and -9 and -18. Now that I have this sort of graph, I can go ahead and try to visualize a five X using these points that I found. So at zero F of X is equal to one. So somewhere around here at one F of X is equal to nine, at two X is equal to 81. So already off the charts, you know, if I were to draw a dot Maybe somewhere up there, but I'll just no, I'll increase it very steeply to. So keep that as a sort of guide And at -1, I got a fraction of 1 9th. So somewhere there and then at -2, I get an even smaller fraction of one over 81. So if I connect all these dots, you can see that exponential shape. And again, at two, it goes up very steeply as we expect for exponential functions. And now that I have F of X, let's build G of X off of the F of X function. So G of X is equal to negative nine to the X. Well realize that nine not to the negative X to the X realize that nine to the X is the function of F of X. So I can say that G of X is equal to negative alfa Becks. So if I do negative F of X is equal to G of X, I just need to multiply by negative one to all of these outputs, which means that when X is equal to zero G of X is equal to one times negative one which is negative one and so on for all of these. And this is also known as a reflection. So that negative sign in front of nine to the X is causing G of X to be reflected across the X axis. So we expect to see a reflection across the X axis, the reflection would be across the Y axis if B negative had been in the exponent here. But because it's on the outside, it's going to be across the X axis. So that means that I expect to see G of X go in the sort of opposite direction of F of X. So this blue line here is what I expect G of X to look like. And this black line is F of X. And you can see that there's that reflection across the X axis and notice also how I just reflected G of X without necessarily plugging in these points that we found. So that's why transformations are helpful because it can quickly help us decipher the shape that we're expecting the graph to look like. And if we look at answer choice, A, you can see that the shape is not the one that we're expecting. So answer choice A gets eliminated and answer choice B has that reflection that we're expecting. So it could be the correct one. And let's just check, see and again, see has a reflection across the Y axis which is not what we're looking for we want across the X axis. So she gets eliminated. So let's go back to be. So V has the correct reflection that we're expecting. So now we can focus on finding the horizontal A cento and the domain and range for these functions. Well, the horizontal ascent is the point where the functions get really close but never achieve that value. And in this case, there's this sort of Asientos Y equals zero. And we can confirm that that Aceto exists because when we look at nine to the X and I plug in in this case, a negative value for X. So X to the negative 1000, I get nine to the negative 1000, which translates to a fraction that is saying one divided by nine To the 1000 power. So although this gets really close to zero, it could never equal zero because it's still a fraction, a really small fraction but still not zero. And if I look at nine or negative nine to the X and I plug in once again, a Really large negative numbers such as negative 1000, I got negative 9 to the negative 1000. And like before this becomes a fraction where now I have the negative in front. So negative 1/9 To the positive 1000. So now you can see that we are still approaching zero with this fraction, but now it's a negative number and that's where we see that reflection but there's still that a cento at Y equals zero. So I can see that I have a horizontal awesome tote Y equals zero. And for the domain of both of these functions, you can see that the domain is the possible X values that the functions can take with the functions still being defined. And because the inputs or the X values are in the exponents, we can plug in any value into the exponent and none of the, none of the outputs will be undefined because of a certain X value that we plug in. So that means that X can be all real numbers. And if we show that an interval notation, we can mark it as negative infinity to positive infinity. And now when we want to look at the range, well, that's the possible why values that the function G of X can take on. And from here, you can see that that's going to be different than F of X because F of X exists solely in the positive realm, but G of X exists solely in the negative realm. So that means that the range for G of X will be from negative infinity up until zero. And we need to do a parentheses because as we saw before It won't ever reach zero but it will get really close. So we'll do a parentheses there. So this becomes my final answer for this problem. And you can see that answer choice B also has the domain listed as negative infinity to positive infinity. The range from negative infinity to zero and the horizontal A cento at Y equals zero. So B is our final answer. Hopefully, this video helped and see you next time.

Key Concepts

Exponential Functions

Transformations of Functions

Asymptotes

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