Graph each function. Give the domain and range. ƒ(x) = (1/3)

Question

Graph each function. Give the domain and range. ƒ(x) = (1/3)^x-2 + 2

Accepted Answer

Hello everybody. I'm doing right today. Today, we're going to put again this question that states provide the domain and range of the function by graphing it. Now, the function is F of X is equal to negative open parentheses, one divided by 11 closed parentheses to the exponent X minus seven and separate from the exponent A plus five. Now for a question like this, I want to determine, I see that our F of X is kind of complicated convoluted. So I want to see if there's any way that I can break this down to its simplest form and go from there. So that will be the parent function here. And I'm going to label that parent function as G O X is equal to open parentheses, one divided by 11, close parentheses to the exponent X. So that will be our most basic function here. This is our parent function. And basically how we get from the G FX to the F of X is just making different translations and transformations. And that's how we get to the FX. So that's exactly what we're gonna do. We're gonna start with our G FX parent function and slowly work our way to, to our F of X and see what changes we have to make. Now for our parent function, it's also important to note that this graph, the G of X will have a horizontal ascent tode at so to have a horizontal asci tote at the X axis, so the X axis here will function as a horizontal asil. So that's something that we have to keep in mind. And with that in mind, let's go and start making the transformations here. So we have a parent function G of X. Now let's leave our first change as H of X and we'll have H of X is equal to and now we'll add the negative seven to the exponent of the minus seven. So we'll have H of X equal to open parentheses, one divided by 11 closed parentheses, X minus seven. Now, and this means that we shift G of X seven units, right? So that's what we do. Once we add that minus seven to the exponent. Next, let's do another change. We'll have K of X now and now we'll add the negative in front of our equation. So we'll have K of X is equal to negative open parentheses, one divided by 11 closed parentheses to the exponent X minus seven. Now, this means that H of X reflects about X axis. So we are reflecting now our graph will reflect now to uh around the X axis. So finally, we can add the plus five at the end of our equation to have that F of X, it's equal to a negative open parentheses, one divided by 11 closed parentheses X minus seven and then plus five at the end. And this means that we shift K of X five units up. And now I can try different points in our F of X equation since that's the one we're going to be working with. So first I'll try X equals seven. And if I have X equal to Z seven, our exponent will be 07 minus 70. So an exponent to the zero power is one. So we'll have negative one plus five is four. So if X equals seven, Y equals four, now we can also try X equal six. Now if we have six in our X, our exponent will be six minus seven, which is negative one and we have negative one in our exponent, our fraction will be flipped. So will have, will be left with negative 11 and negative 11 plus five is negative six. So will have a Y equals negative six or this can be rewritten as the points seven comma four and six comma negative six. Now, with all this information, we can go ahead and graph a general idea of what our graph is going to look like. So will graph alpha is equal to negative open parentheses. One divided by 11 closed parentheses to the exponent X minus seven and outside of all that plus five. Now let's go back to the horizontal Asymptote that we noted. So we noted that we had in a horizontal ascent tode at the X axis. And now we look at all the transformations we made, we see that one of our transformations was moving to the right, that's not gonna affect our, our horizontal as tode we reflect and we move five units up. So moving five units up means our horizontal Asymptote will also be moving five units up because everything will be moving up. So the horizontal ascent or the H A will be now at A Y equal five. So our horizontal as all those changes are horizontal ato will move from the X axis to Y equals five. So now let's draw a graph. So we'll draw a Y axis and an X axis and we all draw a dash at A Y 05 and another one at A Y 04. Since I know one of our points has a Y of four. And then I'll also draw a dash at a Y of negative six. Since I know one of my points will be there, I'll also draw a dash at an X of seven and an X of six. And now I can plug in the points that I obtained. So I know at an X of seven, I'll have a Y of four. So I'll put a point at seven comma four and I'll put another point at negative six, comma or six, comma negative six. So an X of six and A Y of negative six and I can connect those two points. And I know my graph will continue extending downwards to the left. And as we said, we'll have a horizontal Asymptote at A Y of five. So our graph once we pass the 0.7 comma four will be approaching a wild five but never touching it. So we're going to be turning right and always approach that Y of five but never touches and will continue to expand right. So now that will be it for the graph and we can talk about the domain and the range. So when it comes to domain, we are talking about the possible X values in our graph. So here we look at how, how our graph behaves left and right. And we see that our graph extends infinitely left and it extends infinitely right. Meaning our domain is an open parentheses from negative infinity to positive infinity because we go left and right infinitely. Now, for the range, we have to see how our graph moves up and down as in the Y axis. So we're dealing with the possible Y values in our graph. So first we look how our graphic sends downward and we see that our graphic sense infinitely downward. So we'll have an open parentheses from negative infinity two. And now we look at our graph moves upward. And as we mentioned, we have a horizontal atom to look at A Y 05. So our graph goes up until A Y 05 never touches it, but always approaches a Y 05 and we never touch it or pass A Y 05. So that is where our range will lead end up. We won't go past five. So our range will be from a negative infinity to five and that will close the parentheses and it will be a parentheses because we never touch that five. It's a horizontal Asma tilt. If we touch that five, then it would be a bracket, a closing bracket. But because we don't, it's a closing parentheses. And just like that, we have been able to graph our function F of X as well as obtain its domain in range just from looking at that graph. So I hope that was helpful. And until next time.

Graph each function. Give the domain and range. ƒ(x) = (1/3)^x-2 + 2

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Graph each function. Give the domain and range. ƒ(x) = (1/3)x-2 + 2

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Graph each function. Give the domain and range. ƒ(x) = (1/3)^x-2 + 2