2nd Semester Review

追加: 2026-05-19

[00:00:01]Okay, in this review we're going to cover a few key properties of what we have seen in the spring semester. Let's begin with the first question.

[00:00:11]For the first one, it's asking us for the function h of x equals the square root of x, what is the maximum, what is the minimum, and so on. Now, when we graph this on Desmos, let's go ahead and type it in. We have what we can refer to h of x as y, so we have y equals the square root of x. When we graph this, the minimum will always be this point right here. If we click on if we click on it, it's going to be the point 0, 0. So, let's go back here and we're going to write down that the minimum that we have is going to be located at the point 0, 0. Now, the maximum is not going to be possible because our graph is going to continuously go upwards. We will never be able to identify the actual maximum point. The domain for this one will be from 0, including it, comma, all the way to positive infinity. And that's what we can see right there based on the graph.

[00:01:17]We start at the point 0, 0. It's a solid point, so we have to include it with the bracket. And the graph will continuously go towards the right-hand side, and that's why we indicate that the domain is going to be from 0 to positive infinity. Keep in mind that domain is going to be read from left to right.

[00:01:37]Okay? Now, for the range, it is read from the bottom to the top.

[00:01:46]Okay? Now, based on what we just saw from the graph, we're going to indicate that the range began at the point 0 right there for the y value 0. We're going to include it as well. It was a solid point. And the graph will continuously go upwards, so hence we have positive infinity. Okay. So, now you hopefully got the gist of this in terms of graphing this on Desmos. I'm going to go and speed it up a little bit right here. So, for the function f of x equals the negative square root of x plus three minus five, the minimum point is going to be located actually for this one it might potentially be a maximum based on the graph that you have. Uh it is going to be located at -3, -5.

[00:02:32]We will indicate that the domain will be from the following. We Keep in mind that this H value that we have right there, we have to change the sign. So, instead of a positive three, it's now a -3. For the K value, notice that it's a -5, so leave it alone, okay? Do not change the sign. And we could have done the same process for the previous function right there. You can notice that the H value is a zero, the K value's a zero if it's not written down.

[00:03:04]And that's why we had that minimum point right there.

[00:03:07]Okay. So, based on that information, we can indicate that the uh domain for this function is going to be located at -3 to positive infinity.

[00:03:18]And the range is going to be from negative infinity to -5, including the five, right? Excuse me, including the -5. And now we're going to move on to the uh next graph, which is g of x equals x minus four to the third power plus two.

[00:03:42]Now, let's go ahead and graph it. We have x minus four to the third plus two.

[00:03:48]Let's graph it right there.

[00:03:50]So, we have parentheses x - 4 close parentheses to the third power.

[00:03:56]And then we have plus two. Okay? So, now when we graph it, keep in mind that this is not our minimum point right here.

[00:04:05]This point of inflection, this point of inflection right here is the turning point. Okay? 4,2 is the turning point.

[00:04:12]So, it's not a minimum. It's not a maximum. You do not have that for these types of function types. Okay? So, for this one, we're going to indicate that we have none.

[00:04:24]Okay? And we notice that for the domain, it's read from left to right. So, do we have a graph on the left for the x values? Yes, we do. Do we have a graph on the right-hand side for the x values?

[00:04:34]Yes, we do. So, hence we're going to indicate that the um function will have a domain from negative infinity {comma} to positive infinity. Now, for the range, it's the same idea. Keep in mind that we read this from bottom to top. So, we ask ourselves, do we have a graph in the bottom for the y values? Do they continuously going downwards? Yes, they do. Do we have a graph on top? Yes, we do. The y values continue to go upwards.

[00:05:01]So, hence we will indicate that the range will be from negative infinity to positive infinity.

[00:05:08]Whoops.

[00:05:10]Okay, let me fix that real quick.

[00:05:13]Okay, now let's move on to our next function type, which is h of x equals the cube root of x + 2 - 4. Let's go ahead and graph that on our calculator.

[00:05:25]So, we have let's see the cube root. We can type it in as well, CBRT, just like that. And we have x uh plus two.

[00:05:39]And then outside, we have minus four.

[00:05:43]Like that.

[00:05:45]Okay, we're going to zoom out a little bit. We can see that the graph is down below. Okay, and now keep in mind that for this one we do not have a minimum, we do not have a maximum, but we do have a point of inflection. If you see right here, this turning point, let me see if I can actually uh Uh well, I can't click on it. Well, we have a turning point right here, a point of inflection. So, we do not have a minimum, we do not have a maximum. So, based on that information, let's go ahead and write it down that we do not have a minimum or maximum for this one.

[00:06:20]And for the domain, remember it's read from left to right. So, do you have X values on the left-hand side? Yes, you do. Do you have uh X values on the right-hand side? Does the graph continue? Yes, it does. So, hence we will indicate that the domain will be from negative infinity to positive infinity. Now, for the range, remember it's read from bottom to top. So, do we have a graph in the bottom? Yes, we do.

[00:06:42]So, that means that the Y values are going to negative infinity. Do we have a graph on top? Yes, we do. So, that means that the Y values are going to positive infinity. So, hence we have a range that's read from negative infinity to positive infinity.

[00:06:58]Okay, awesome. So, just keep that in mind if you ever want to identify the uh domain, range, minimum, maximum, vertex, and so on, you want to graph them on Desmos and remember that domain is read from left to right and range is read from bottom to top. Okay, and keep in mind that if you have a solid point, it's going to be read with the bracket.

[00:07:20]Okay, I just want to remind you. So, if you have a solid point, it's a bracket.

[00:07:24]If you have an open circle, it's always going to be a parenthesis right there.

[00:07:29]Okay, so just keep that in mind.

[00:07:31]Okay, for the next question, we have if two graphs are inverses of each other, which statement is true? Okay. Now, uh Uh, I will tell you up front that we have the following for the uh inverse functions. Let's go ahead and write.

[00:07:51]Um, let's see. Let's see.

[00:07:54]Okay. So, we have let's see this y equals the square root of x.

[00:08:04]And for this one, let's go ahead and type in y equals x squared.

[00:08:12]Okay.

[00:08:13]Uh, but we want to minimize, let's see if I recall correctly how to do that.

[00:08:19]All right. We're going to indicate that x has to be greater than Oh.

[00:08:25]or equal to zero.

[00:08:28]There we go. All right. So, right here, we can clearly see that uh that these two functions are inverses of each other. These square root of square root of x and the quadratic function x squared are inverses of each other. Now, we know that they're inverses of each other because we can see right here that we have what? A reflection across the line y equals x. Let's Let me go ahead and graph that so that you guys can see that. Y equals x. So, if you ever see symmetry across this line y equals x, that means that the functions are inverses of each other. So, keep that in mind. That's a very, very important detail.

[00:09:08]Uh, so we're going to indicate that this statement is true and we're just going to keep track of it right there.

[00:09:13]Okay. And then it says that the x values and the y values are interchange. And that is true. Whenever we are looking for the inverses of each other, we are always switching the x and the y values.

[00:09:24]The domain and the range values interchange as well. So, what does this mean for us? It means that all the statements above are true, okay, for inverses of each other. Now, down below for question number three, we have the following. For the parent square root function y equals the square root of x, state the following.

[00:09:44]Okay, for this one, let's go ahead and graph the square root function right there.

[00:09:50]Uh we can see that the graph begins at the x value zero, the y value zero, okay? So, based on that, we can answer some of these questions right here. We can indicate that the domain is going to begin from zero, including it, right? It's a solid point, to positive infinity. The range, remember that's from bottom to top, right? We didn't have a graph in the bottom, but we did have a y value that started at zero, and the range continued to go upwards. So, hence, we have a positive infinity for the range at the very end. For the x-intercept, we saw that the x-intercept is going to be at the point zero {comma} zero. The minimum point is going to be at zero {comma} zero as well.

[00:10:39]Okay? And for the y-intercept, we all Luckily for us, for this one, we had a y-intercept at zero {comma} zero, as well. And for the maximum point, we do not have one because the graph will continue to go upwards. You can clearly see that on the graph. We will never have an x at y value. If we just keep zooming out, zooming out, you can see that the graph will continuously go upwards. We will never have an x at maximum value.

[00:11:06]Okay. Now, for the next one, we have for the parent cube root function y equals the cube root of x, state the following.

[00:11:13]Now, for this one, let's graph it on Desmos.

[00:11:17]Uh keep in mind that the shortcut uh keys are what? CBRT, okay? You can also click on the functions button right there.

[00:11:25]Now, if we graph this, our graph is going to look like this. Kind kind looks like a snake shape. And we can answer the following questions. From this one, we can indicate what that the domain is going to be from negative infinity to positive infinity, right? The graph was going to the left-hand side and to the right-hand side. So, that's why we indicated that. And for the range, our graph was going from what? Bottom? Did we have graph in the bottom? Yes, we did. Did we have a graph in the top?

[00:11:51]Yes, we did. And both of them continuously went up and and to the bottom. So, hence the range is going to be from negative infinity {comma} to positive infinity.

[00:12:00]Then, for the Y intercept, we had a Y intercept, if we look at the graph that we saw right here, we have a Y intercept at 0 {comma} 0, and that will also be the X intercept, as well. So, let's write that down. 0 {comma} 0 for the X intercept, as well as the Y intercept.

[00:12:18]Now, I want to add some other information right here. I also want to tell you guys that just looking at this graph right here, we can see that we have symmetry across the origin, the point 0 {comma} 0, right? So, for this one, we're also going to write down this important piece of information that we have symmetry about the origin.

[00:12:54]Okay?

[00:12:55]And we also saw right there that we do not have a minimum nor a maximum. So, that's a very important detail for this one that we do not have a minimum nor a maximum. So, let's put a little star right there and write down no minimum, no maximum. Okay? [clears throat] So, keep that in mind.

[00:13:14]Okay. Now, down below for number five, they're asking us to determine the type of function graph, then state the domain and the range for each one. Well, looking at this graph right here, notice that it's going to be a square root of function. Now, it's not easy to identify it, but we can kind of see that it's a flipped upside down graph. So, for example, if I went to Desmos and I were to type in y = negative square root of x, you can see how it's upside down, right?

[00:13:45]Uh so, that's how I can tell the difference, okay? All right. So, for this one, um we're going to write down that this is a square root function.

[00:13:59]And for the domain, we're going to see what that the vertex right here started at the point -2, 4.

[00:14:08]negative 4.

[00:14:10]So, based on that, for the domain, we're going to indicate what? That we're including the point -2 for the x value, and that it's going to go to positive infinity, right? The graph continues to go to the right-hand side.

[00:14:24]And for the range, it's read from bottom to top. So, do we have a graph in the bottom? Yes, we do. So, that's going to be referred to as what? Negative infinity. Do we have a graph on top?

[00:14:34]Yes, we do. Or excuse me, no, we do not.

[00:14:37]We have a stopping y value. Remember, we're looking at the y value for the range at -4. And we're going to include it because it's a solid point. So, let's go ahead and put that bracket right there.

[00:14:51]Okay. Now, let's go on to the second graph right here. For the second graph, notice that this is going to be a cube root function. So, let's write that down that this is a cube root function.

[00:15:08]And we're looking at the point of inflection right there. So, just keep that in mind, that point is going to be located at -4, positive 2.

[00:15:17]And we can see that domain is read from left to right. So, do we have a graph on the left? Yes, we do. Do we have a graph on the right? Yes, we do. So, hence we have from negative infinity to positive infinity for the domain. For the range, do we have a graph in the bottom? Yes, we do. It's very slow to go to the bottom, but eventually it's going to go to the bottom.

[00:15:35]And that's going to be referred to as what? Negative infinity. And do we have a graph on top? Yes, we do. So, it's going to go to positive infinity for the range. And now for part C. For part C, notice that the graph is in its example of a cubic function. Let me write that down. A cubic function.

[00:16:03]My apologies right there. Cubic function. Okay. And now for the domain.

[00:16:08]From left to right, we can see that we have a graph on the left, so that's going to be negative infinity. Do we have a graph on the right? Yes, we do.

[00:16:14]So, that's positive infinity. And for the range, do we have a graph in the bottom? Yes, we do. Do we have a graph on top? Yes, we do. So, hence from negative infinity to positive infinity.

[00:16:25]Okay. Now, to number six. Okay? Take a look at this. They're asking us to solve for each equation. So, let's go ahead and solve for the first one. My first step is going to be to subtract three on both sides. Like this.

[00:16:42]This will leave me with the square root of X is equal to -3. I'm going to bring the work up here.

[00:16:50]So, I have the square root of X is equal to -3.

[00:16:55]I will then square both sides. Like this.

[00:17:00]This will leave me with X is equal to uh positive nine.

[00:17:05]And that's going to be my final solution for part A.

[00:17:08]Now, for part B, for this one, I will subtract nine on both sides. Like this.

[00:17:17]And I will bring the work up here.

[00:17:22]Okay? So, for this one, I now have what?

[00:17:26]The square root of 4x + 8 is equal to positive 2.

[00:17:33]I will then square both sides like this.

[00:17:37]And this will leave me with 4x + 8 is equal to positive 4.

[00:17:43]I will then isolate for x by subtracting 8 on both sides. This will now leave me with 4x is equal to -4. I will then divide both sides by 4.

[00:17:54]And this will leave me with the final solution as x is equal to -1. Let's go ahead and box that in.

[00:18:02]Okay, let's move on to the next one, uh part C.

[00:18:06]Now, for part C, this is a cube root function, right? So, just keep that in mind. All right, from there, my first step is to get rid of that cube root.

[00:18:14]How do I do that? Well, I will go ahead and cube it, okay? So, the opposite of the cube root is to cube. So, we're going to cube or to place to the third power each side.

[00:18:25]That will get rid of the cube root, and I am left with x -2 is equal to what? 2 to the third gives me positive 8. I will then add 2 on both sides.

[00:18:38]This will leave me with x is equal to positive 10 for my final solution.

[00:18:43]Now, for problem D, my first step is to do what? To add 11 on both sides like this.

[00:18:53]I will end up with the square root of x + 4 is equal to -3 + 11 is going to leave me with a positive 8.

[00:19:10]Okay? And then I will cube both sides like this.

[00:19:16]And this will leave me with the final solution as X + 4 is equal to what? 8 to the third power is going to leave me, let me grab my calculator real quick.

[00:19:34]That's going to give me 512.

[00:19:36]So, let's write that down. 512.

[00:19:39]And then I will subtract four on both sides.

[00:19:43]And that will leave me with X is equal to 508 for the final solution.

[00:19:53]Okay, and now let's move on to problem E. For problem E, my first step is to cube both sides like this.

[00:20:04]I will then end uh then end up with 3X + 1 is equal to 64.

[00:20:11]Then I will subtract one on both sides.

[00:20:14]This will leave me with what? 3X is equal to 63.

[00:20:18]I will then divide both sides by three.

[00:20:21]Right?

[00:20:24]Let me type that in. 63 / 3. Okay. And that's going to leave me with X is equal to 21.

[00:20:36]Okay.

[00:20:37]All right, let's move on to the next one for number seven.

[00:20:41]Okay, so now for number seven, they're asking me for the following. Find the inverse of each function and list any domain restrictions, okay? Now, we're not going to do all of them. I just want to get to the core idea for each, okay?

[00:20:54]So, if I look at the first one, I am given a table of values and they're asking me to find the inverse, right?

[00:21:01]Well, we know that for the inverse, we just simply switch the the and the range values with each other. We interchange them.

[00:21:09]Or essentially, the domain and the range are interchanged. So, if I make my new inverse function, my new table of values down below, I now have the new X and the new Y. So, I switched them. So, now I have 5 {comma} 3, 2 {comma} -1, and -4 {comma} +6 for my final solution.

[00:21:32]Okay?

[00:21:33]Now, for problem B, they're asking me for the inverse, okay? So, here's my first step. If I want to find the inverse, I'm going to rewrite f(x) to be y, so I have y = the square root of x + 8 - 11. I will then switch x and y. So, now I have what? x is equal to the square root of y + 8 - 11 outside. Then, I will solve for the new x by isolating the new y.

[00:22:09]Uh excuse me, to find the new inverse by isolating the new y. So, I will add 11 on both sides like this.

[00:22:20]Okay, let me bring the work up here.

[00:22:22]And this will now leave me with x + 11 is equal to the square root of y + 8. I will then get rid of the square root by squaring both sides like this.

[00:22:35]And this will leave me with {parentheses} x + 11 to the second power is equal to y + 8. Now, do not expand it. Leave it the way that it is so far.

[00:22:47]And we're going to subtract 8 on both sides. And again, these are not like terms, so do not expand them. Do not put them together just yet. You're going to leave them separated. So, we now have what? x + 11 to the second power minus eight is equal to Y.

[00:23:05]And that's going to be the new inverse function. So, now when I rewrite it, we're not going to do the domain restriction, so don't worry about that.

[00:23:12]So, when I rewrite it, I will end up with parentheses X plus 11 to the second power minus eight.

[00:23:25]And we can check that this is true because if we were to graph them, we would see the uh reflection across the line Y equals X. So, first, let's go ahead and sketch that right out.

[00:23:38]So, let me see.

[00:23:40]I have the square root of X plus eight um minus 11 and then the inverse that we found, which was Y equals You don't have to type in the proper notation inverse. Um we just want to see whether it's true or not. And we had what? Parentheses X Um let me go back real quick. Plus 11 close parentheses to the second power and then minus eight.

[00:24:15]Yeah, minus eight.

[00:24:17]Okay.

[00:24:18]And keep in mind that we did have Well, you're not going to You're not going to do the restriction, but for this one, we will indicate that the restriction is going to be X equals um negative 11.

[00:24:35]There we go.

[00:24:52]Okay, let's see.

[00:25:00]>> So, x = 11.

[00:25:07]No, x cannot be that value.

[00:25:12]Let's see.

[00:25:14]Where's the slash?

[00:25:19]Okay. Well, I I can't show it right here, but it's only going to be the right-hand side of the graph, and we will see the reflection across the line y = x, okay? So, if we just focused on the right side of the parabola right there, we can see that we do have the um we do we do have the symmetry right there, okay?

[00:25:41]All right. So, now let's move on to part C.

[00:25:45]Okay, and it's pretty much the same idea right here. We're going to go ahead and rewrite f of x to be y, so let's go ahead and show that. We have y = We then have x + 1 to the second power minus 5. And then we're going to switch x and y, so now we have x is equal to parentheses y + 1 to the second power minus 5. Then we're going to solve for the new y by isolating for it, so we're going to add 5 on both sides like this.

[00:26:18]We're going to bring the work up here.

[00:26:20]Okay. So, now we have x + 5 is equal to y + 1 to the second power.

[00:26:28]Then we're going to take the square root. For this one it's a little bit different. You have to take the square root, right? That's the opposite and how we get rid of the squares, the exponents. Um like this, okay? So, now this is going to end up leaving us with the square root of x + 5 is equal to y + 1.

[00:26:48]And then we subtract 1 on both sides.

[00:26:51]Keep in mind that you want to leave them separated. They're not like terms, so do not combine the five and the minus one.

[00:26:56]And this is going to leave you with the square root of x + 5 is equal or excuse me, minus one is equal to y.

[00:27:06]Or essentially for the inverse, we just found it. It's going to be the inverse of f of x is equal to the square root of x + 5 minus one for the final solution.

[00:27:17]Okay? All right. So again, we're not focused on the restrictions. Uh we're just focused on those steps. So just remember that for the inverses, we rewrite f of x, we switch x and y, and we solve for the new y. That's the process, okay?

[00:27:33]All right. So let's go down below.

[00:27:35]Okay, and this is just a reminder for number 11. It says to solve the equation the square root of x + 5 equals two.

[00:27:43]Jerome entered uh into Desmos into the graphing uh portion of Desmos. And what's the solution according to this?

[00:27:50]Okay. Well, keep in mind that if you are ever looking to solve for an equation, let's say the say that we want to solve for the square root of x + 5 is equal to two, you can always type it into Desmos in that format. You can see right there that the left-hand side of the equation right there was typed in as y equals and the right-hand side of the equation, the two right there was typed in as y equals that uh value.

[00:28:19]And what you're going to look for is the point of intersection. That's how we solve it. So just in case uh and we're going to do those problems by hand, but just in case you're stuck, "Hey, how do I solve this?" with the graphing calculator, that's what you do. You're going to type in the left-hand side of the equation right there, y equals, and the right-hand side of the equation, y equals, and you're going to look for the point of intersection. In this case, we see that the point of intersection is located at -1, {comma} +2. So that means what? That the solution is x = -1.

[00:28:54]That's the solution right there.

[00:28:56]Okay? So, just keep that in mind that you can always do this with the graphing calculator.

[00:29:02]All right. So, now we're going to move on to number 13, okay? Now, for number 13, they're asking us to complete the table below by identifying the asymptotes, domain, and range of the given function. Keep in mind that you can type this into Desmos, uh yourself.

[00:29:22]And you're going to end up with the following solutions, okay? So, if we look right here, without the graphing calculator, this, for the exponential function, will always be my horizontal asymptote. So, y = 1 right there. That's my HA, my horizontal asymptote, okay? Uh From there, let's go ahead and write that down.

[00:29:46]So, we have y = 1.

[00:29:49]And for the domain, let's type it in on Desmos, right there.

[00:30:00]I want the full thing.

[00:30:04]Okay.

[00:30:05]Uh We're going to type in y = You can also type in the w of x, uh two to the power of x - 6.

[00:30:18]Now, when you type it in, it's not going to let you, so you need to wrap it in parentheses, okay? So, we have parentheses x - 6, like that.

[00:30:27]Then we click on the right arrow key, we type in plus one.

[00:30:31]And now we can actually see the graph, okay? And you can see right there that the graph will get very close to that y = 1, but it will never actually intersect it. It's going to get to the very small decimal values.

[00:30:44]Uh and we can see that the graph right there has a domain that's going all the way to the left, all the way to the right-hand side. And uh for the range, it's not going to intersect y = 1.

[00:30:57]But it will increase to um as many uh y values that are above y = 1. So, for the domain, we're going to indicate that it's going to be from negative infinity {comma} to positive infinity. For the range, it's going to be from one, not including it because it's the asymptote, {comma} all the way to positive infinity.

[00:31:19]Okay?

[00:31:20]And now for the um we're not going to do part uh B, this one. Don't worry about that one.

[00:31:28]Let's focus on B of X, okay? So, for this one, we have seen it before in our first unit for the rational functions.

[00:31:34]This value right here is going to represent our vertical asymptote, okay?

[00:31:39]We just changed the sign. So, this is going to turn into what? X uh equals -8. That's the vertical asymptote. And this value that's outside is my horizontal asymptote. So, we have HA is equal or, excuse me, has y = 5, okay? Now, keep in mind that we do have a reflection across the x-axis, but that's not going to affect the domain, that's not going to affect the range for this type of function.

[00:32:09]So, let's indicate that right there. We just indicated the vertical asymptote HA as well.

[00:32:14]For the domain, it's going to be from negative infinity {comma} it's based on the vertical asymptote, so it's going to be from negative infinity to -8 union from uh -8 {comma} to positive infinity. And keep in mind that these are vertical asymptotes or these are asymptotes overall, so you do not include them with the brackets. They have to be with the parentheses.

[00:32:38]Okay, and then for the range, it's going to be from negative infinity, comma, to five.

[00:32:47]Union from five, comma, to positive infinity. Keep in mind that it's based on the horizontal Y value.

[00:32:54]Okay, so now let's move on for the next one.

[00:32:59]Okay.

[00:33:01]Let me get my documents.

[00:33:05]Okay, so now for number 14, it says answer the following questions for the function f of x equals 3 to the power x + 1 represented in the table.

[00:33:16]It says, what is the asymptote of the function? Well, if you recall from the previous example that we just saw, this K value represents the horizontal asymptote that we have for the exponential function. So, we have an HA at Y equals 1. That's our solution right there. So, HA, horizontal asymptote at Y equals 1.

[00:33:37]And you can clearly see that from the table of values, we will never get exactly at Y equals 1. We're going to approach it, we're going to get very close, but never actually be uh intersecting that Y value.

[00:33:49]Okay, now it's then asking us what are the domain and range of the function.

[00:33:54]Okay, so for this one, we're going to indicate this is We're going to indicate that the domain is going to be from negative infinity, comma, to positive infinity. And again, this is based on you graphing it. So, if you graph it on Desmos, you're going to see that.

[00:34:12]And for the range, we're going to indicate that this is going to be from um It's going to be from negative one, not including it, to positive infinity.

[00:34:29]Okay?

[00:34:30]Just like that. Okay? All right, and then it says circle the value in the table showing the solution when f of x equals 82. Essentially, we're looking for what? When y equals 82, what's the x value? Well, we can clearly see that right here at x equals 4. That is our solution.

[00:34:50]And we just circled it on the table of values. We also have the precise solution x equals 4.

[00:34:55]Then it's asking us what's the inverse of the function, okay? Now for this one, let's go ahead and solve it. So we have what? Uh we change f of x to be y, so we have y equals Let me fix that.

[00:35:15]I'm sorry, I'm not sure why the pen is doing that.

[00:35:19]Okay, so we have y equals uh 3 to the power of x plus 1, okay? Now I am going to switch x and y, so now I have x equals 3 to the power of y plus 1. Then I will isolate for y by subtracting 1 on both sides. This will now leave me with x minus 1 is equal to 3 to the power of y.

[00:35:53]Then I will convert this to a logarithm.

[00:35:56]Keep in mind that the inverse of the exponential function is a logarithm. So we have log base 3 of um of x minus 1 is equal to y.

[00:36:18]Okay, and that's the correct format right there. That's going to be our final solution for the inverse.

[00:36:27]Okay. All right, So, now let's move on to number Number 15 is a good one.

[00:36:35]Okay, for number 15, it says the population of the town NumberVille is currently 12,500 people and declining at rate of 3,000 per Excuse me, 3% per year.

[00:36:48]Okay, it's asking us write a function that can be used to determine the population P of T of NumberVille in T years.

[00:36:56]Uh so, for this one they give you the or excuse me, let's put it in the precise format. So, we know that the function type is always going to be uh be given to us in this format, Y equals A times B to the power of T. Okay, that's the format that we're looking for. So, based on that, we can see that the A value, which is the initial value, is going to be 12,500.

[00:37:23]The R value right there is going to be 3%, but keep in mind that we need to convert that to be what? A decimal value. Okay, let me erase this real quick, this part here. To be more precise, it should be um A times 1 plus or minus R to the power of T. That's the more precise format.

[00:37:49]Okay, so if we look at this, the percentage is going to turn into what?

[00:37:54]Point zero three. Okay, so now if I put all of this together and keep in mind that it says declining, right? So, if it's declining, it's going to be subtraction.

[00:38:05]Now, if it said growth or that it's growing, that it's appreciating in value anything of that nature, you would put a plus. Okay, so just keep that in mind.

[00:38:15]So, based on that we will write down the following equation. We have Y equals 12,500 right there times one plus the point zero three Let me zoom in further.

[00:38:38]Okay. point zero three Okay, to the power of t.

[00:38:47]Okay, and let's put that together fully.

[00:38:50]So, if we put that fully together, that's going to turn into y equals 12,500 Why is it doing that?

[00:39:06]Okay.

[00:39:08]There we go. 1.03 to the power of t. There we go, and that's our final solution right there.

[00:39:16]Okay.

[00:39:17]Now, it's then asking us use your function to approximate the population in five years. Okay, so what does that mean for us? It means that t is equal to five, and we're going to substitute that into that equation right there. So, our equation, when we do the substitution, is going to be y equals, or excuse me, p of t, right?

[00:39:38]Because that's what they're asking us.

[00:39:40]Or p of five now.

[00:39:42]Okay, is equal to 12,500 times 1.03 to the power of five. Okay, and if we type that Oh, actually, now that I think about it, we did say declining, right? So, this should be subtraction. My apologies about that. Let me erase that real quick. So, that would change things up right there.

[00:40:08]Okay, and that's going to turn into what? 0.97.

[00:40:12]My apologies for that.

[00:40:14]Okay, and then right here, it's going to be the same thing, 0.97.

[00:40:21]Okay, and then we're going to say that the solution is going to be what? If we type it into the calculator, um this should give us um 10,734 people.

[00:40:39]Okay?

[00:40:40]And it makes sense, right? The population is decreasing, so over 5 years we will have 10,734 people. And that does make sense right there.

[00:40:50]Okay, we're not going to do part C.

[00:40:51]We're going to scratch that one off. And then it says, "In how many years will the population become half of its current size?" Okay? So, what what does half mean? Well, it means that instead of 12,500, we're going to divide that by two, and that's going to leave us with 625, correct?

[00:41:10]So, from there, that's the value that we're looking for. So, essentially, we want our function 12,500 uh * 0.97 ^ T to be exactly at 625 uh people. That's what we want. Okay?

[00:41:33]Right there. Or 6,250. I'm not sure why I put that.

[00:41:38]There we go. And there we go. All right.

[00:41:42]So, now from there, here's how we're going to type it into Desmos. We're going to type this as the first line, okay, on Desmos, this as the second line on Desmos, and keep in mind that for each one, you're going to type in Y equals, for this one Y equals as well.

[00:41:58]Okay, let's go ahead and show that on Desmos. So, if we type it in, let's see.

[00:42:06]12,500 (.97) ^ T.

[00:42:14]Uh we can say X instead of T, your choice. And then we have Y = 6,250.

[00:42:20]Okay, and then we do what? We look for the point of intersection. Keep in mind that these numbers are very large, so when you zoom out, make sure that you zoom out as much as you can so that you can see the values.

[00:42:32]Okay, and what we're going to do is click on the point of intersection.

[00:42:35]Right there, I can see that the X value or the T value is going to be what? Approximately 22.75.

[00:42:44]Uh let's see, is it in years? Yeah, in years, okay? So, it's going to take approximately 22, 23 years for the population to be cut in half, okay? So, let's go back here and we're going to indicate that it's going to be if we round it to the nearest tenths place, okay? We're going to say that T is going to be approximately uh 22.

[00:43:12]8 years. Okay, for the population to be cut in half.

[00:43:17]All right, so let's see another example of this idea. Let's move on to number 16.

[00:43:25]Uh for number 16, we have the following.

[00:43:28]It says the population of a certain uh town can be modeled by the function P of N equals 245,000, 1.03 to the power of N, where N represents the number of years since 2020. Is the population increasing or decreasing?

[00:43:43]Well, we can see right here that we have a value that's being added to the one, right? The original one. Remember, the format should be what? Y equals the A value, which is the initial, uh then times one plus or minus uh the R value to the power of T, correct? That's the original format. So, based on that, we can see that something was added to the one. So, hence this is going to be an increasing population.

[00:44:12]Let's go ahead and circle that right there.

[00:44:15]And then it's asking us, "What is the percent increase from year to year?"

[00:44:18]Well, we can see right there that if we were to take out one from that, right?

[00:44:23]Minus one, we're going to be left with.03 or essentially 3%, correct? So, that's our solution right there, 3%. And it then says, "Approximately when would you expect the population of the town to be 400,000 people?" So, we're going to take this function that we see above, right? The 245, 000 * 1 Hold on.

[00:44:55]* the 1.03 to the power of n, and we're going to set it equal to the 400,000.

[00:45:03]That's what we're looking for, okay? So, on Desmos, this is going to be the first line, and you're going to type in y = this. This is going to be the second line. You're going to type in y = that, and you're going to look for the point of intersection. Let's go ahead and look for it ourselves. So, if we go to Desmos and we type it in, let's see.

[00:45:25]We have uh 245, 000 parentheses 1.03 to the power of x, and then we have 400,000.

[00:45:46]There we go.

[00:45:49]And again, this number's going to be very large, so let's zoom out.

[00:45:56]All right, there we go.

[00:46:00]All right. And that's going to be approximately 16.6 years if we round to the nearest uh the nearest tenths place uh when the population will reach uh will reach 400,000 uh people, okay? So, for this one, we're going to say what approximately 16 What did we say before?

[00:46:22]16.6?

[00:46:24]Okay, so we have n is equal to approximately 16.6 years for the solution.

[00:46:36]Okay, now for the next one. Now, this is a very good question. Uh so, this one says what? Mr. Moneybags has a million dollars he wants to give away. So, that's the initial value, right? That's the A value, right? A equals 1 million dollars.

[00:46:56]Okay? And excuse me.

[00:46:59]Each day he will uh give away half of the money to different uh to a different charity, right?

[00:47:04]So, the R value, right? It's going to be R equals half means what? 50%, right?

[00:47:13]So, we have 50%, but if we convert that to a decimal, we have what? 0.5, correct? And it says, "How many days will it take before he is down to $15.26?"

[00:47:25]Okay, now we're given a table of values, but it's still very difficult for us to see that, right? So, we're going to write down the function ourselves. So, we have for the function Y equals a million for the initial value, and we're going to do what? Since it's declining, right?

[00:47:45]He's giving away the money, it's going to be 1 minus the 0.5, which is going to leave us with positive 0.5. So, that's going to be our rate, 0.5 to the power of t, where t represents the number of days, correct? And we want it to get to what? $15.26.

[00:48:04]So, we're going to take this equation that we just found $1 million for the initial value.

[00:48:10]Okay.

[00:48:15]And we're going to set it equal to uh $15.26.

[00:48:21]So, on Desmos, you're going to type in this. For the first part, you're going to type in y equals and you're going to type in this for the second part as y equals and you're going to look for the point of intersection. Let's go ahead and type it in ourselves. So, we go to Desmos. We're going to type in $1 million.

[00:48:47]And then that's what? Uh.5 close parentheses to the power of x and then that's going to turn into $15.26.

[00:49:02]Okay. So, for this one, we want to scroll down because the number's going to be very small.

[00:49:11]All right. There we go.

[00:49:17]All right. So, now if we click at the click on the point of intersection, uh we're going to get uh roughly 16 days, right? So, and again, we're thinking about whole numbers, right? Uh in this case scenario. So, the person, Mr. Money Bags, is going to go from a million dollars if he's given half of the money every single day, in 16 days, he's going to run out of money and he'll be all the way down to 15. Uh 15.26 dollars, right? So, let's go ahead and write that down that the final solution t is going to be 16 days. Okay? That's the final solution. You're going to see something like this.

[00:49:57]Okay. Now, let's move on to the next one for number 18 part D. We're not doing every single thing, okay? So, it says write the equation in exponential form. Now, if you look right here, notice that we have log of X is equal to 3. So, when the base is not written down, it is always known to be a 10. Okay? Now, when I convert this into the exponential form, I now have what? 10 to the power of 3 is equal to X. And that's my final solution. Just simply that.

[00:50:32]Okay?

[00:50:33]Questions?

[00:50:35]Nope. Okay. Well. All right, let's move on to number 19. Don't overthink think that one. That one's very simple. For number 19, we're going to do this question right here. It's asking us to solve for X. So, my first step is to convert this to the exponential form.

[00:50:51]So, I take my base and I place it to the power of X is equal to 27.

[00:50:58]Right? That's in the correct format.

[00:51:01]And now, we're asking ourselves 3 to the power of X. What what X value will give me 27, right?

[00:51:08]And you can check with the Desmos calculator. You can do it by yourself or just based on what you know. In this case, we know that 3 to the power of 3 is equal to 27, right? So, hence X is equal to 3. And that's going to be your final solution. Now, keep in mind that we already showed you the format. Let's say that you want to solve for this. You You forgot every single thing. On Desmos, you're going to type in the left-hand side of the equation.

[00:51:33]Like this.

[00:51:34]You're going to type in log. Right, we have to click on the functions.

[00:51:42]Scroll down right here to the calculus section.

[00:51:46]And we have log base um three of 27, right?

[00:51:54]And that's going to give you three right there. But, you can also do the right-hand side, which is what y = 3 like that. And if you zoom out, you're going to see the point of intersection between the two.

[00:52:07]Let's see what happened here.

[00:52:09]y = There we go.

[00:52:14]Well, it's the same value, right? So, they're going to be very close to each other uh right there. Well, we're just going to click that, right?

[00:52:22]Okay. Well, this one didn't work out work out nicely. Uh but, that's the format that you do this by hand. Uh let's go down to problem I. Uh and again, for problem I, if the base is not given to you, it is known to be a 10.

[00:52:36]So, we then convert this to the exponential form. We have 10 to the power of two is equal to 5x - 11. We know that 10 to the second power is 100.

[00:52:49]So, we have 5x - 11. We then add 11 on both sides.

[00:52:57]And this is going to be what? 111 is equal to 5x. We then divide both sides by five like this.

[00:53:07]And this will leave us with a final solution as x is equal to 111 / 5. If you want to convert that to a decimal, you're more than welcome to do so. But, that's the format that we saw for this equation. Okay? Now, let's move on to the next one.

[00:53:24]Uh problem G, okay? And again, if you recall what I said prior to this, I told you guys that you can type uh you can separate the work. You can say that this is going to be y = on the left-hand side. And then on the second line of Desmos, you have this as the y = and let's type it out. Hopefully for this one, it works out nicer on Desmos.

[00:53:47]Okay, let's type it in.

[00:53:50]So, we have 9 to the power of And keep in mind, you want to wrap this in parentheses. If you have more than one term, wrap it in parentheses. So, we have x - 6 close parentheses.

[00:54:03]And then we have y equals We have what?

[00:54:07]Parentheses 1/3.

[00:54:11]And then close parentheses to the power of x. There we go.

[00:54:17]And what we're looking for is the point of intersection. So, where do we see them intersecting? Well, we see them intersecting at x = 4. So, when x = 4, that's going to be the solution that will make them both equal to each other.

[00:54:30]Okay? Nevertheless, this is what we can do to solve it by hand. Again, we know that the solution is going to be x = 4.

[00:54:37]Okay? But nevertheless, let's do it by hand.

[00:54:40]Uh so, for this one, we have what? 9 to the power of x - 6, right?

[00:54:46]Or we can convert that to be 3 to the second power, right? 3 to the second power gives me 9. And then we have x - 6.

[00:54:57]And on the right-hand side, we can convert this to be what? Parentheses 3 to the power of -1, right? 3 to the power of -1 gives you 1/3, okay? And then x outside.

[00:55:08]And then we do what? We simplify. So, we have what? 2 * x gives us 2x. 2 * -6 gives us -12.

[00:55:17]And that's equal to what? 3 times Excuse me, 3 to the power of -1x.

[00:55:25]Okay? And when the bases are the same, you can set the exponents to be equal to each other. So, now we have what? 2x - 12 is equal to -1x.

[00:55:36]And for this one, you can do it in so many different formats. Let's say that we move uh all the x's to the right-hand side. Let's say that's the case scenario.

[00:55:45]We do what? -2x.

[00:55:50]And that's going to turn into what? -12.

[00:55:52]The two x's will cancel out. = -3x.

[00:55:56]We then divide both sides by -3 like that.

[00:56:00]And this will give me what? 4 is equal to x, or essentially x = 4, which is exactly what we found right there.

[00:56:08]Okay, and here's uh the next one. Uh for this one, same idea. Remember what I said before, you can type it into Desmos. So, we have what? y = You type that in. For this one, we have y = as well for the second line.

[00:56:22]Okay, let's type it in. Let's see what we end up with.

[00:56:26]Um so, we have y = 16 to the power of {open parentheses} x + 1 {close parentheses} and then we have y = {open parentheses} 1/2.

[00:56:53]All right, and then to the power of 3x.

[00:56:57]Okay, there we go.

[00:56:59]All right, and then we're going to zoom out, zoom in, do what you have to do, and we're going to look for the point of intersection. So, where do they intersect?

[00:57:07]And they intersect at this particular decimal value. So, for this one, we want to be very precise. Let's see that this is number uh 19 part H.

[00:57:20]Yeah, 19 part H.

[00:57:24]Yeah, for this one, we're going to get a decimal value. We can also convert it to a fraction. Okay, so let's put it together. All right, so we have what? 2 to the power of what will give me 16?

[00:57:35]Well, it's going to be two to the power of four.

[00:57:38]And we then have the original exponent x + 1 equals on the right-hand side. We have two to the power of -1.

[00:57:48]And then originally we have the exponent 3x and now we simplify. So we have what two to the power of what 4 * x gives us 4x 4 * 1 gives us positive 4 equals -1 * positive 3x gives us two to the power of -3x.

[00:58:06]And now when the exponents are the same we set the excuse me when the bases are the same we set the exponents to be equal to each other. So we have 4x + 4 equals -3x.

[00:58:18]We will then move all the variables to one side all the constants to the other side. It's up to you how you go about doing so.

[00:58:26]So in this case let's move this 4x to the other side.

[00:58:30]And that's going to leave us with 4 equals -7x. We divide both sides by -7.

[00:58:39]And this is going to leave us with a final solution of x equals -4 over 7.

[00:58:45]Right? Or if you type that into Desmos remember that we got -0.57.

[00:58:51]If you go to Desmos if you type that in -4 / 7 you can see it's the same value -0.57 right? So it's the same idea right there -0.57.

[00:59:04]Okay, there we go. That's our final solution.

[00:59:09]All right, so now let's move on to the next part. Let's see.

[00:59:14]What's a good one?

[00:59:17]Okay, for this one number 21 going to change things up a little bit right here. Hopefully you change it on your paper if you have the space.

[00:59:28]Okay, I'm just going to erase these right there.

[00:59:43]Okay. So now, for this one, let's say that we are asked, of the functions listed below, which have the same domain? And let's say that for this one, we have it for Oops.

[00:59:57]Uh for f of x is equal to, let's say log base 3 of x, okay? Let's say that's what we're looking for. We're looking for the same domain as this one. And we have the following options. Let's say that the first option, let's call it number one, right there.

[01:00:17]Uh we have log base 3 of x + 4.

[01:00:22]For option two, we have uh log base 3 of x uh + 5 outside. Option three, we have log base 3 of x.

[01:00:39]Um Oh, let me fix that. Um ch ch ch ch ch ch ch ch ch ch 2 + log base 3 of x, okay?

[01:00:50]And finally, the last option, let's say that we have 4 * log base 3 of x. So, for for this one, the one that we have in red, the domain for it will be from zero and including zero, right? So, if we were to graph it on Desmos, we would go from zero to positive infinity, right? Okay? Now, we then go to check for the graph and the the of the first one. If you were to graph it, uh you would see that the domain for this one would be from -4 {comma} to positive infinity.

[01:01:29]And for the second option, even though it was vertically shifted upwards, that will not affect the domain value. It's still going to go from 0 to positive infinity.

[01:01:40]Same idea with the one down below. They moved the two to the front, but that's still a vertical shift upwards. So, hence, it will still remain as a domain from 0 to positive infinity.

[01:01:52]And for the last option, we have a vertical stretch by four units. Now, keep in mind that the vertical stretches will not affect the domain of the logarithm. So, for this one, we still have the domain from 0 to positive infinity. It's only the inside that does the shifting for the domain to the left, to the right, based on the operation that we see. So, which of them has the same domain? It would be option two, three, and four. Which of them does not have the same domain would be option one. Okay? So, just keep that in mind.

[01:02:23]You might potentially see something like that in the future.

[01:02:27]Okay?

[01:02:28]All right. So, now for number 22.

[01:02:33]Okay? For number 22, we're looking at 22.

[01:02:38]Okay. For this one, uh the domain of the function is all real numbers. And we have the following functions right there. Now, I graphed them beforehand, so I will tell you what the solutions are. I got false, true, true, and true.

[01:02:57]Uh for the next one, I got false, true, true, false.

[01:03:04]For the next one, I got false, uh true, true, false.

[01:03:12]Uh for the next one, I got true, false, false, true.

[01:03:24]I'm sorry, not true. That one should be false.

[01:03:27]Okay, and then for the next one I got true, true, true, and false.

[01:03:34]And for the last one I got false, false, false, and true. Okay?

[01:03:40]All right, and again this is all based on just simply graphing your functions on Desmos.

[01:03:46]Okay, so now let's go down below.

[01:03:48]Uh for this one, uh we're not going to do part A because that's not a nice one.

[01:03:53]Uh let's look at number 26. For number 26 it says uh what? Rewrite each function in its equivalent graphing form, then give the equation for the asymptote. So for this one, uh we are going to do uh long division.

[01:04:09]So we're going to take the top in, bottom out, right? So we have 3x + 7, and then we have the x - 2 outside. And then uh we ask ourselves 3x / x, right?

[01:04:23]The first term divided by the first term is going to give us what? It's going to give us three. We then take this three, and we multiply it to both terms outside. So 3 * x gives us 3x, 3 * -2 gives us -6. We then wrap that in parentheses, place a subtraction sign, and we distribute the negative symbol.

[01:04:42]So instead of a positive 3x, we now have a -3x. Instead of a -6, we now have a positive six when I distribute the negative symbol. Okay? So now we can scratch that off, and now we put all of it together. So 3x - 3x is going to cancel out. Seven um plus six is going to leave you with 13.

[01:05:02]And there's nothing else that I can do.

[01:05:03]So this 13 is going to be my remainder.

[01:05:05]Okay? So now I'm going to place it in the graphical format, so I have y equals the remainder 13 divided by what we were dividing by x minus 2 and what we were able to divide by which was the plus 3.

[01:05:21]So, you're going to see something like this on the exam.

[01:05:25]Okay.

[01:05:26]All right, and then let's move on.

[01:05:30]Let's see.

[01:05:39]Okay, number 30. Okay, I'm going to change something up right here.

[01:05:45]I'm going to add to this question.

[01:05:46]Number 30 uh right here. We're going to skip that first part.

[01:05:52]Okay, so for this one it says Romero's Banquets charge one fee for the facility and guest. Anna it seats 40 people to attend her party and she calculates the cost per person at Romero's will be $20 for each person, right? So, what is the total fee for uh at Romero's? Okay, so we're going to calculate the total cost.

[01:06:15]We have 40 people.

[01:06:17]Each person will pay $20, right?

[01:06:20]So, the total will be $800.

[01:06:25]That's the total amount that they will pay.

[01:06:28]But now, let's add to this question, okay? This is what I want you to add.

[01:06:32]What if 20 peop- uh 25 people uh go instead.

[01:06:44]Okay, so instead of 40 people, we now have 25 people and we still have to share the cost, right? The total cost is $800, right? All together that's what we know the total cost is. That's what we were able to um uh calculate, right? So, instead of 40 people, now it's 25 people. So, what do we do? We take the $800 and we're going to divide it by 25. So now that's going to give us the individual cost for each person, which will be $32 per person.

[01:07:19]Okay, instead of the 40 people that show up.

[01:07:23]Okay, and now finally, here's the last one, number 31. This is all based on transformation, okay?

[01:07:31]Uh so we're given the functions right there, but they're asking us to do the transformations for them. Okay, so here's the first one. I'm going to focus on the first one right there.

[01:07:41]For y equals the square root of x. So for this one, it says that we're translated to the left five units.

[01:07:49]So for this one, we have y equals the square root of x plus five, okay? Then it says what, translation up n units, right? So that's going to be what, y equals the square root of x and then outside we have plus n.

[01:08:06]And then it says reflected over the x-axis, so that's going to be what, y equals negative square root of x, right?

[01:08:12]The negative symbol has to be outside.

[01:08:14]And then it says what, translated six units down and two units to the right.

[01:08:18]So for this one, we have y equals uh the square root of x minus two, right? It's going to the right and then minus six outside. There we go.

[01:08:33]And expanded vertically by a factor five, you need to have that number outside and in front of the function, okay? Let's move on to the next one. For the next one, we have y equals x to the third power and it says translated up n units. So for this one, we have y equals n uh excuse me, x to the third power plus n. For the next one, we have y equals what, negative x to the third. Y equals for the next one, we have parentheses. Now for this one, if you ever have more than one thing or something that happens to the left or the right, we have X Uh and for this one, it's what? To the right, so we have minus two to the third power and then minus six outside.

[01:09:20]All right, there we go. And then finally, expanded vertically by a factor of five, so we have Y equals 5 X to the third power like that.

[01:09:30]Okay, for the next one, let me change the color up so that we don't get confused. Okay, so for this one, we are translating to the left five units, so we have Y equals the cube root of X plus five.

[01:09:45]Okay, we then have Y equals the square root of X plus N, then we have Y equals negative the cube root of X, then we have Y equals the cube root.

[01:10:06]Uh of X uh minus two and then minus six.

[01:10:14]Right. And then we have Y equals 5 times the cube root of X. There we go.

[01:10:21]Okay, and then for the next one, we have uh let's see. Y equals uh two to the power of X plus five. Y equals negative two to the power of X.

[01:10:45]Then we have Y equals two to the power of X minus two minus six down below and then Y equals five and then parentheses 2 to the power of X like that.

[01:11:03]All right, and then for the next one we have Y equals log of X + 5.

[01:11:14]Then we have Y equals log of X + N.

[01:11:22]Then we have Y equals log of X - 2 and then - 6.

[01:11:36]And then Y equals 5 log of X.

[01:11:43]All right, and then last one we have Y equals 1 over X + 5.

[01:11:52]Y equals 1 over X + N.

[01:11:57]Y equals 1 over X - 2 - 6. And then finally Y equals 5 over X.