Graph each function. Give the domain and range. See Example 3. ƒ(...

Question

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

Accepted Answer

Hello everybody. I'm doing all right. Today, today we're going to begin this question that states provide the domain and range of the function by graphing it where our function is EFAX is equal to negative open parentheses. And then a fraction one divided by 11 closed parentheses to the exponent X plus six and then outside of the exponent minus three. Now we're looking at a question like this. First, I'm gonna ask myself is what's the parent function here? Because I see that our FVE function seems relatively complicated. And is there any way that I can break this down to its most basic form and then go from there? So my parent function here, I'm going to label it G O X and it's going to be one divided by 11 to the exponent X. So that's going to be our parent function. And it's important to note that that function has a horizontal asym tote or an H A at the X axis. So now our mission is to see how we get from this parent function G of X to our function in the question F of X. So let's go little by little until we get there. So we had G of X is equal to one divided by 11 to the exponent X. So now let's make another function. This one will be H of X H of X is equal to one divided by 11. And that will make a small change. We'll say that we had open parentheses, one divided by 11, closed parentheses. And now we add the plus six to the exponent. So the exponent will be X plus six now. And this means that we shift G of X six units left. So that's what happens when you add the plus six, the G FX graph moves six units to the left. So now let's add another step in this question and we'll label another function, let's say K of X, so K of X and now we'll add the negative in front of everything. So we'll have negative open parentheses, one divided by 11 closed parentheses to the exponent X plus six. And this means that our H of X graph reflects about the X axis. So now we've added the plus six to the exponent and the negative in front. And we see how our graphs have been affected. Next. All we're missing now is the minus three at the end of our function. So we'll go back to Afro backs now and we'll have that equal to negative open parentheses, one divided by 11 closed parentheses to the exponent X plus six and outside of the exponent with a regular function minus three. And this means that K of X shifts three units down. So now we went from our parent function G of X is equal to one divided by 11 to the exponent X to our current function little by little. So we know we have to move that parent function will be shifted six units left reflected about the X axis and then shifted three units down. So now let's plug in some points. So let's say we have H is equal to negative six. So we, when we have that H is equal to negative six, we'll have a zero in our exponent and anything to the exponent zero is one. So we'll have a Y of negative four because at the end, we'll have negative one minus three which is negative four or a point at negative six comma negative four. Now we can try another numbers or another X will say X is equal to negative seven and then we have negative seven as our X will have a negative one in our exponents. And once we continue to do that math or we know a negative one is gonna make the reciprocal of our uh fraction and will be left with negative 11 minus three, which is going to be A Y equals negative 14 or the point negative seven comma negative 14. So now with all of that information, we can graph our function so will graph F of X is equal to negative open parentheses, one divided by 11, close parentheses to the exponent X plus six and outside of the exponent with our normal function A minus three. Now let's keep in mind the horizontal as to that we had with our parent function, we know our parent function G of X had a horizontal ascent to at the X axis. Now, when we look at all the changes that we made, we saw that one of our changes was reflecting about the X axis and then we shifted three units down. So now that horizontal Asymptote will be at a Y of negative three. So we'll have horizontal ask him to at Y equals negative three. So now we can go ahead and continue to graph this. So we'll draw a Y axis and we draw an X axis like just like that. And now we can plug in different points that we had. So we know we're gonna have a point at an X of negative six because that's one of the Xes we plugged in. So we know we're gonna have a point at negative six and another point at an X of negative seven and the point of negative six will intersect or will have a point at A Y of negative four. And the point of negative seven will have a Y, a point at A Y of negative 14. And as we said, we know we're going to have and horizontal Asymptote at A Y of negative three. So we can connect those two dots and continue. And once we pass the point at negative six comma negative four, we have to turn to the right as we approach that the Y of negative three but never touch it. And the graph will continue forever towards the right now, we can talk about our domain and our range. So our domain will be an open parentheses. And now we look at the graph and we see that our graph extends infinitely to the left and infinitely to the right, which means our domain will be negative infinity to positive infinity. Now remember that a domain has to do with the possible X values in our graph. So if our graph extends infinitely to the left and to the right left and right have to do with our X values. Therefore, we extend from negative infinity to positive infinity. And finally, we look at the range, the range on the other hand has to do with the possible Y values of our graph. So we have to see how it moves up and down the Y axes. So we see that as we move to the left, we're also extending infinitely downwards. So we go to negative infinity that would be comma and then as we move to the right, in our graph, we see that we're going up and we have a horizontal Asymptote at A Y of negative three. So we approach A Y of negative three but never touch it and we never go past that Y of negative three. So our graphic stems from a range of negative infinity to negative three and close that parentheses. And just like that, we've drawn a graph for our function F of X and we've also determined the domain and the range for set function. So I hope that was helpful. And until next time.

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

Key Concepts

Graphing Functions

Domain and Range

Exponential Functions

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