AP Precalculus Unit 1 Review

In this video, I'm going to review unit 1, AP Precalculus, unit 1, part 3, covering topics 112, 113, and 114. All of those are topics that we're going to revisit in all of our units, so these are really important. Starting with topic 112, Transformations of Functions. So I'm going to go over the different transformations, and then we'll do some... problems. Okay, so I have a document here. You can find this in the description for the different transformations. So in general, and you're going to start seeing this more and more, especially as we get towards like trig functions and things like that, but in general, if we have g of x equals a f of b x plus c plus d. So the a, b, c, and d are going to transform our graphs. First I'm going to talk about the additive transformations. So additive transformations are going to be translations. So first we're going to look at if you add c directly to our x, so on the inside. Anything that directly affects the x is going to affect our graph horizontally. So if we have this, that's going to be a horizontal translation of negative c. That's how you're going to see College Board describe a transformation from this. So we have two options, either c is greater than 0 or c is less than 0. So if our c is greater than 0, a little bit counterintuitive, our graph is going to move to the left. So here I have my original function f. This is f of x plus 2. The whole graph, all of the x values get shifted to the left too. versus if what's on the inside is negative, it's going to go to the right. So I have my same graph of f, but now I've transformed it with an x minus 2, and it moves it to the right. If I add directly to the function, we're affecting the y values. So this is going to be described as a vertical translation of d. If d is greater than 0, we're going to go up 2 units. So I have my f of x. And my f of x plus 2 after the parentheses moves up 2. If it's negative, it's going to go down 2. So here's my f of x. This is f of x minus 2. Next, I have my multiplicative transformation. So this is when we're multiplying the function either on the outside or the inside. So if we're multiplying on the outside, just like if we're adding on the outside, we're affecting the graph vertically. So we're going to have a vertical dilation by a factor of a. Graphically, we're looking at two cases. Either that number is bigger than 1 or that number is really between 0 and 1. We'll talk about what happens if it's negative. If it's bigger than 1, we're going to have a vertical stretch. We're going to multiply all the y values by that number. So I have f of x and then my second graph, the dotted one, is 2 f of x. So you can see that all the y values get stretched, they get pulled away from the x-axis by a factor of 2. So this point that had a y value of 1, negative 1 has a y value of negative 2. Here we were at negative 3, it gets pulled all the way down to negative 6. If it's less than 1, or between 0 and 1, my graph is going to get compressed vertically. So it's going to squish it towards the x-axis. So the same kind of points I pointed out, this one that had a y value of negative 1 now has a y value of negative 1 half. This one that had a y value gets multiplied down. So negative 1 half is down now at negative 3. So it just gets stretched, or sorry, compressed. It gets squished that way. Last we have our second multiplicative. We're on the inside, and the inside just doesn't behave nicely. But it's a horizontal dilation by a factor of 1 over b. So if I have a graph that has a, if that value is bigger than 1, then it's going to compress the graph horizontally. So if I had a factor of 2, that means I have a horizontal dilation by a factor of 1 half. So all of these x values get squished into the y-axis. So this x value of 3 is now an x value of 1 half. Versus if that number is less than 1, like 1 half, it's going to get shorter. stretched horizontally. So here I've got my f of x, and we have f of 1 half x, which means my factor is 2, so all my x values get doubled. So this one that was at 3 is now at 6, and so on. So it's stretched horizontally. Okay, next are my reflections. If we have the opposite, of f of x, it's going to be reflected in the x-axis. So the whole thing, if we had a y value that was positive, is going to become negative. If we have a negative, it's going to become positive. The whole thing just gets flipped upside down. Versus if we have a negative on the inside, it's going to reflect it over the y-axis. So a positive x value becomes a negative x value. A negative x value becomes a positive x value. So those are all the different transformations that you need to know. So let's look at some examples. Okay, so my first task, I have a graph of f of x, and then we're going to graph g of x using whatever transformations the equation tells us we need to do. So first we're going to describe the transformation, and then we're going to graph the graph. Now on your AP exam you're not actually going to be physically graphing, but I think it's great practice in actually physically seeing what's happening. Alright, so our first graph that we're looking at is g of x is f of x plus 2. So this is on the inside. If it's on the inside, it's going to affect the graph horizontally. Okay, so how we're going to describe this, this is a horizontal translation of negative 2. It's always going to be the opposite. I'm also putting in parentheses because I think it's a little bit more descriptive that we're going left two units when we're doing this transformation. So I'm going to take my points and move them to the left. So I'm going to take that one, move it left, take that one, move it left, take that one, move it left, take this one, move it left, and then connect our dots. This next one, we have f of x minus 2 on the outside. If it's on the outside, it affects the graph vertically. Okay, so this is going to be a vertical translation of negative 2. So physically we're moving the graph down 2. So we're going to move all these points down 2 units. Shape of the graph doesn't change, just its location. In this one, we have two different transformations. We're going to treat the negative and the two separate. For the negative, that's going to be specifically a reflection in the x-axis. Then we have the two, which is going to be a vertical dilation. So we have a vertical dilation by a factor of 2, which is going to stretch the graph vertically. So we've got two different things. I'm going to do the negative first and then stretch it, or you could stretch it and then do the negative. So here I'm going to, instead of being at negative, one I'm going to go to positive one and then double it to two. Here this is going to go from negative two to positive two and then doubled to four. Here this is going to go from a positive two to a negative two and then doubled. Here this is going to go from a negative one to a positive one and then double it. Okay, so it's reflected and then stretched, or you can stretch it and then reflect it. Both will result in the same graph. Okay, here we're on the inside, which is going to affect the graph horizontally. This will be a horizontal dilation. So we're going to have a horizontal dilation by a factor specifically of 2, the reciprocal whatever's here, which is going to stretch our graph horizontally. So it's going to take all of our x values and pull them away from the y-axis. So instead of being at negative 2, this is going to be negative 2. This will have an x value of negative 4. Negative 1 is going to get pulled away to negative 2. 2, or sorry, this is at 1, is going to get pulled to 2. And this 4 is going to get pulled all the way here to 8. You get a graph that just gets stretched out. Okay, here I have two additive transformations. So my two transformations here is a horizontal translation of 1, where it's going to go to the right. When this is a minus, we go to the right. And then we've got a vertical translation up 2. So I'm going to take each of these, go right 1, up 2, right 1, up 2. Rate 1 of 2, rate 1 of 2. The last one we're going to graph is f of negative x, which is going to reflect the graph in the y-axis. So we're going to take all of the x values and make them opposite. So this is going to flip over to positive 2, this is going to flip over to positive 1. This is going to flip over to negative 1. This is going to go to negative 4. So there's my reflection in the y-axis. Next we're going to identify a transformation and write the resulting equation. So we have the same graph, and then I'm just moving it. So if I look at it, here I've got a point at negative 3, 1. That corresponding point is at negative 1, negative 2. So it has moved 2 to the right and down 3. And those are translations. So for my equation, g of x is going to be f of, if I'm moving to the right 2, we're going to have an x minus 2 on the inside, down 3 is going to be minus 3 on the outside. Okay, next one here. This whole thing is squished horizontally. So this is going to be, and if you look at it specifically, like negative 3 is now at negative 1.5. This 3 is at 1.5. Okay, so we have a horizontal dilation by a factor of 1 half. So g of x is equal to f of, it's always going to be the reciprocal. So it's going to be 2x is what's going to give me that horizontal dilation. Here, the whole thing gets flipped over the x-axis. Okay, so if that's the case, g of x is going to equal the opposite of f of x. It makes all the y values opposite of what they were. Okay, now rather than seeing the transformations on a graph, we're just explicitly told what they are. And again, we're going to write an equation. So first we're going to write an equation. If our graph is dilated vertically by a factor of 5, dilated horizontally by a factor of 1 third, translated up 10 units, and to the left 9 units. Or you might see it say translated negative 9 units. So my g of x. Okay, vertical dilation of 5 means I'm going to have a 5 in front of my f, dilated horizontally by a factor of 1 third. So my b value is going to be a 3, and then we're going to put the translation in parentheses. So we're translated to the left 9, so I'm going to have x plus 9 on the inside, and then it's translated up 10, will give me a plus 10 there. Okay, similar thing here. Dilated horizontally by a factor of 8, dilated vertically by a factor of 3 fourths, translated left 5 units, and down 23. So my a value is going to be whatever my vertical translation is. If the horizontal dilation is a factor of 8, then my b value is going to be 1 eighth. If we're going to the left, this again is going to be a plus. And then if we're going down, we're going to subtract that 23 on the outside. Next it says, describe the transformations in correct order to construct of. F to construct the graph of G. So we're identifying the different pieces. So here we have to deal with the 6, the 10, and the negative 5. As far as order, the multiplicatives need to happen before the additives, but you can do multiplicative vertical, additive vertical, followed by multiplicative horizontal, additive horizontal. I'd like to do the multiplicative ones first, and then do the translations. So my first multiplicative is a vertical dilation by a factor of 6. My other multiplicative is the horizontal dilation of a factor of 1 tenth. And our additive is the minus 5 on the inside, which means we have a horizontal translation of 5, which would mean the graph would go 5 units to the right. Let's take a look at our next one. We have a vertical dilation by a factor of 5. That 1 third on the inside is going to give us a horizontal dilation by a factor of 3. And then we have a vertical translation of minus 8, moving the graph down 8. So you might be tested on these transformations from a graph, from the equation. Here we're going to look at what happens to a table of values if we have these translations. So my first one here, we have a table of values for... f of x, and then we're going to do a table of values for g of x with the corresponding transformations. So this one, we just have a minus 5 on the inside, which means it's moving to the right 5 units. So my f of x values will not change. My input values are going to change. So if we're moving all of these to the right 5, I'm going to add 5 to each of these numbers. So negative 2 plus 5 is 3. 0 plus 5 is 5, 4 plus 5 is 9, and 8 plus 5 is 13. Here, I'm going to the left 3 and down 2. So if I'm going left 3, I'm going to subtract 3 from all of my x's. Negative 5, negative 3, 1 and 5. Down 2, we're going to subtract 2 from all my y values. So this will be at 3, 2, 1, and 0. Here I've got two things. This negative is going to be a reflection in the y-axis, sorry, x-axis, which will affect the y values. This 2 is a horizontal dilation. by a factor of one half. That will affect the x values. So for the horizontal dilation of 1 half, that means these values are all going to get cut in half. So this will be a negative 1, 0 will be 0, 4 is going to become 2, 8 is going to become 4. For the reflection in the x-axis, all of these are going to become opposite. Okay, next, rather than going from equation to table, we're going to go from table to equation. So we're going to kind of try to pick up on what's happening going from here to here. So looking at my x values, all of these are two bigger than this, so we're going right to. All my y values are opposite, so it's a reflection in the x-axis. So my equation for g is going to be a negative f of x minus 2, since it's going to the right. Okay, here my y values are staying the same, but my x values are getting multiplied by 1 half. So my g of x is going to equal f of 2x to get that horizontal dilation. Okay, next, transformations. Obviously, if we're... stretching and moving and translating graphs that can affect the domain and range. So we have a function that's domain goes from negative 8 to 4 and the range goes from negative 2 to 9 and we're going to find the domain and range of of the resulting graph. So here I've got g of x is f of x minus 2 plus 3, so we're going right 2 and up 3. So my domain will be affected by this, my range will be affected like this. So if my domain is negative 8, 4, and I'm going to go to the right 2. We're going to add 2 to these. So my domain is going to be negative 6, 6. My range of negative 2, 9, if we're going up 3, we are going to add 3 to these. So negative 2 plus 3 is 1. 9 plus 3 is 12. Okay, here we have a reflection in the x-axis. We have a vertical dilation by a factor of 2. And then we're going to the left one and down three. Okay, so I'm just going to do this one step at a time. We're going to look at what's happening to the domain versus what's happening to the range. So domain is going to be affected by the reflection, the vertical dilation, and the vertical shift. So reflection in the x-axis is going to take these... Sorry, these are going to affect my range. So this is going to be a reflection, which is going to make this negative 2 become a positive 2, and this 9 to become a negative 9. But really, that just changes... my range to be going from a negative 9 up to a positive 2. My vertical dilation is going to multiply those values by 2. So now I'm going to be at negative 18 to 4. And then we're going to move it down 2, or down 3. So this will be at negative 21. and 1. The only thing that's affecting the domain is this left one, which means I'm going to subtract 1 from both of those. So that was a lot of questions on topic 112. That is a topic that you will see over and over and over again. So it's one you really want to get down. Now we're going to look at 113 and 114, which is going to be all about modeling different situations. So my first situation is an open top box is to be made from a 30 by 18 rectangular piece of paper. or a piece of cardboard, by cutting equal squares of length x from each corner. So we're going to take a little chunk out of each corner. And then fold those up, which is going to make our open-topped box. So we're told, write an equation for v of x, the volume, and what is the appropriate domain? Okay, well my volume is going to be length times width times height. I'm going to start with the height because when I fold these guys up, my height is going to be x. My length is going to be this dimension here. Well, if this whole thing is 30 and I'm taking x out of here and x out of here, this is going to be 30 minus 2x. And then similarly here, this is going to be 18 minus 2x. Because same thing, I'm taking an x chunk out here, an x chunk out here, so that's going to be 18 minus 2x. The domain is my set of x values. So obviously I can't have an x value of like 100. I can't have an x value of negative 10. Those don't make sense in this problem. So at a minimum... My x has to be bigger than 0. Okay, at my maximum, and really that's going to be dictated by this side, the biggest chunk of x I could possibly take out would be 9 from both sides, but I'm just going to make it less than 9, because if it is 9, then we just kind of basically cut the whole thing in half. But if it's a little bit less than 9, we'll have a little bit of the box left, so that would be my domain. Okay, next model we're going to make. An ice skating rink charges $10 for two hours of skate time. Each additional hour costs $3 to skate. Write a piecewise equation that gives the cost C to skate for T hours. Okay, so this piecewise is going to have two equations. One for the first two hours and one for the additional hours. So we have a flat fee of $10. When we have been skating for 0 to 2 inclusive. If we stop at 2 hours, we're good with $10. After that, we're going to get charged an additional $3. But we're only going to get charged the $3 for each hour after the initial 2. So for my equation, we're going to have... I have 3 for every t value bigger than 2. So for example, if I skated for 3 hours, I'd only want to add one $3 additional charge. So that's where this comes in. This will be when t is greater than 2. We're not told any restrictions on the cap. Maybe you can just keep skating forever. But with our information given, this would be our piecewise. For our last couple of questions, we will be using the calculator to do some models using the regression features of our calculator. I'm going to demonstrate this on the TI-84. If you are using a different calculator, then you'll have to figure out how to do that with those calculators. And remember that Desmos is now going to be embedded during the AP exam, so if you learn how to do the regressions in Desmos, you can use that. Using the TI-84. So we're going to do a quadratic regression and use that to find or predict f of 20. So on the TI-84, you're going to start by pressing the start and we're going to go to edit. If you have data in here, you can highlight the list and do clear enter and that'll clear it. Next, you want to populate your lists. Now that we've populated the list, we want to do the quadratic regression. So you're going to go... Okay, sorry. I can read. Quadratic regression, we nailed it, stored it in y1. So if you go to y1, you're going to see the unrounded answer. I'm going to go ahead and just take these variables and plug them in. I'm going to do three decimals, but we have it stored in our y equals. Okay, so taking those values and plugging them into the equation gives me that. That's not as accurate as we want it to be. So when I evaluate for f of 20, we're going to go to y1, and we're going to evaluate at 20 because we stored our equation in 20. So if we use open parentheses, that will evaluate at the thing I want. So according to this, my f of 20 is approximately 751. Next we're going to do this a cubic regression for the second set of data. So again, stat, edit. We want to get rid of that so we're going to do clear, enter, clear, enter, and then enter our data. If you ever get that error, dim mismatch, what it means is that your lists are not of the same length. So I accidentally had too many, so we're good. So good thing to know what those error codes mean. Now it's happy. So we copy that down, and then we want to evaluate this one for 13. So I'll do Y1, evaluate it at 13. So accurate to three decimals, we get 34.387. So our last example, the number of zebras in a certain area are growing due to conservation efforts. The number of, so the table gives the number of zebras since 2018. The number of zebras can be modeled using a quadratic function. So we're going to write it in the form a t squared plus b t plus c. So I've already entered my data. So I'll do a stack calc, quadratic regression again, store it, copy those values down. And lastly, we're going to use this to predict the number of zebras in the year 2024, which is equal to D equals 5. So we'll do Y1 because we stored it, evaluated it at 5. So we get 82.2, or I would round in the context would be 82 zebras. So this was AP Precalculus Topics 112, 113, and 114. Thank you for watching.

You can find this in the description for the different transformations. So in general, and you're going to start seeing this more and more, especially as we get towards like trig functions and things like that, but in general, if we have g of x equals a f of b x plus c plus d. So the a, b, c, and d are going to transform our graphs.

First I'm going to talk about the additive transformations. So additive transformations are going to be translations. So first we're going to look at if you add c directly to our x, so on the inside. Anything that directly affects the x is going to affect our graph horizontally. So if we have this, that's going to be a horizontal translation of negative c.

That's how you're going to see College Board describe a transformation from this. So we have two options, either c is greater than 0 or c is less than 0. So if our c is greater than 0, a little bit counterintuitive, our graph is going to move to the left. So here I have my original function f. This is f of x plus 2. The whole graph, all of the x values get shifted to the left too.

versus if what's on the inside is negative, it's going to go to the right. So I have my same graph of f, but now I've transformed it with an x minus 2, and it moves it to the right. If I add directly to the function, we're affecting the y values. So this is going to be described as a vertical translation of d.

If d is greater than 0, we're going to go up 2 units. So I have my f of x. And my f of x plus 2 after the parentheses moves up 2. If it's negative, it's going to go down 2. So here's my f of x.

This is f of x minus 2. Next, I have my multiplicative transformation. So this is when we're multiplying the function either on the outside or the inside. So if we're multiplying on the outside, just like if we're adding on the outside, we're affecting the graph vertically.

So we're going to have a vertical dilation by a factor of a. Graphically, we're looking at two cases. Either that number is bigger than 1 or that number is really between 0 and 1. We'll talk about what happens if it's negative.

If it's bigger than 1, we're going to have a vertical stretch. We're going to multiply all the y values by that number. So I have f of x and then my second graph, the dotted one, is 2 f of x. So you can see that all the y values get stretched, they get pulled away from the x-axis by a factor of 2. So this point that had a y value of 1, negative 1 has a y value of negative 2. Here we were at negative 3, it gets pulled all the way down to negative 6. If it's less than 1, or between 0 and 1, my graph is going to get compressed vertically. So it's going to squish it towards the x-axis.

So the same kind of points I pointed out, this one that had a y value of negative 1 now has a y value of negative 1 half. This one that had a y value gets multiplied down. So negative 1 half is down now at negative 3. So it just gets stretched, or sorry, compressed. It gets squished that way. Last we have our second multiplicative.

We're on the inside, and the inside just doesn't behave nicely. But it's a horizontal dilation by a factor of 1 over b. So if I have a graph that has a, if that value is bigger than 1, then it's going to compress the graph horizontally.

So if I had a factor of 2, that means I have a horizontal dilation by a factor of 1 half. So all of these x values get squished into the y-axis. So this x value of 3 is now an x value of 1 half.

Versus if that number is less than 1, like 1 half, it's going to get shorter. stretched horizontally. So here I've got my f of x, and we have f of 1 half x, which means my factor is 2, so all my x values get doubled.

So this one that was at 3 is now at 6, and so on. So it's stretched horizontally. Okay, next are my reflections.

If we have the opposite, of f of x, it's going to be reflected in the x-axis. So the whole thing, if we had a y value that was positive, is going to become negative. If we have a negative, it's going to become positive. The whole thing just gets flipped upside down. Versus if we have a negative on the inside, it's going to reflect it over the y-axis.

So a positive x value becomes a negative x value. A negative x value becomes a positive x value. So those are all the different transformations that you need to know. So let's look at some examples. Okay, so my first task, I have a graph of f of x, and then we're going to graph g of x using whatever transformations the equation tells us we need to do.

So first we're going to describe the transformation, and then we're going to graph the graph. Now on your AP exam you're not actually going to be physically graphing, but I think it's great practice in actually physically seeing what's happening. Alright, so our first graph that we're looking at is g of x is f of x plus 2. So this is on the inside. If it's on the inside, it's going to affect the graph horizontally.

Okay, so how we're going to describe this, this is a horizontal translation of negative 2. It's always going to be the opposite. I'm also putting in parentheses because I think it's a little bit more descriptive that we're going left two units when we're doing this transformation. So I'm going to take my points and move them to the left.

So I'm going to take that one, move it left, take that one, move it left, take that one, move it left, take this one, move it left, and then connect our dots. This next one, we have f of x minus 2 on the outside. If it's on the outside, it affects the graph vertically.

Okay, so this is going to be a vertical translation of negative 2. So physically we're moving the graph down 2. So we're going to move all these points down 2 units. Shape of the graph doesn't change, just its location. In this one, we have two different transformations. We're going to treat the negative and the two separate.

For the negative, that's going to be specifically a reflection in the x-axis. Then we have the two, which is going to be a vertical dilation. So we have a vertical dilation by a factor of 2, which is going to stretch the graph vertically. So we've got two different things.

I'm going to do the negative first and then stretch it, or you could stretch it and then do the negative. So here I'm going to, instead of being at negative, one I'm going to go to positive one and then double it to two. Here this is going to go from negative two to positive two and then doubled to four. Here this is going to go from a positive two to a negative two and then doubled. Here this is going to go from a negative one to a positive one and then double it.

Okay, so it's reflected and then stretched, or you can stretch it and then reflect it. Both will result in the same graph. Okay, here we're on the inside, which is going to affect the graph horizontally. This will be a horizontal dilation. So we're going to have a horizontal dilation by a factor specifically of 2, the reciprocal whatever's here, which is going to stretch our graph horizontally.

So it's going to take all of our x values and pull them away from the y-axis. So instead of being at negative 2, this is going to be negative 2. This will have an x value of negative 4. Negative 1 is going to get pulled away to negative 2. 2, or sorry, this is at 1, is going to get pulled to 2. And this 4 is going to get pulled all the way here to 8. You get a graph that just gets stretched out. Okay, here I have two additive transformations.

So my two transformations here is a horizontal translation of 1, where it's going to go to the right. When this is a minus, we go to the right. And then we've got a vertical translation up 2. So I'm going to take each of these, go right 1, up 2, right 1, up 2. Rate 1 of 2, rate 1 of 2. The last one we're going to graph is f of negative x, which is going to reflect the graph in the y-axis. So we're going to take all of the x values and make them opposite.

So this is going to flip over to positive 2, this is going to flip over to positive 1. This is going to flip over to negative 1. This is going to go to negative 4. So there's my reflection in the y-axis. Next we're going to identify a transformation and write the resulting equation. So we have the same graph, and then I'm just moving it.

So if I look at it, here I've got a point at negative 3, 1. That corresponding point is at negative 1, negative 2. So it has moved 2 to the right and down 3. And those are translations. So for my equation, g of x is going to be f of, if I'm moving to the right 2, we're going to have an x minus 2 on the inside, down 3 is going to be minus 3 on the outside. Okay, next one here. This whole thing is squished horizontally.

So this is going to be, and if you look at it specifically, like negative 3 is now at negative 1.5. This 3 is at 1.5. Okay, so we have a horizontal dilation by a factor of 1 half. So g of x is equal to f of, it's always going to be the reciprocal. So it's going to be 2x is what's going to give me that horizontal dilation.

Here, the whole thing gets flipped over the x-axis. Okay, so if that's the case, g of x is going to equal the opposite of f of x. It makes all the y values opposite of what they were. Okay, now rather than seeing the transformations on a graph, we're just explicitly told what they are.

And again, we're going to write an equation. So first we're going to write an equation. If our graph is dilated vertically by a factor of 5, dilated horizontally by a factor of 1 third, translated up 10 units, and to the left 9 units.

Or you might see it say translated negative 9 units. So my g of x. Okay, vertical dilation of 5 means I'm going to have a 5 in front of my f, dilated horizontally by a factor of 1 third. So my b value is going to be a 3, and then we're going to put the translation in parentheses. So we're translated to the left 9, so I'm going to have x plus 9 on the inside, and then it's translated up 10, will give me a plus 10 there.

Okay, similar thing here. Dilated horizontally by a factor of 8, dilated vertically by a factor of 3 fourths, translated left 5 units, and down 23. So my a value is going to be whatever my vertical translation is. If the horizontal dilation is a factor of 8, then my b value is going to be 1 eighth.

If we're going to the left, this again is going to be a plus. And then if we're going down, we're going to subtract that 23 on the outside. Next it says, describe the transformations in correct order to construct of. F to construct the graph of G.

So we're identifying the different pieces. So here we have to deal with the 6, the 10, and the negative 5. As far as order, the multiplicatives need to happen before the additives, but you can do multiplicative vertical, additive vertical, followed by multiplicative horizontal, additive horizontal. I'd like to do the multiplicative ones first, and then do the translations. So my first multiplicative is a vertical dilation by a factor of 6. My other multiplicative is the horizontal dilation of a factor of 1 tenth.

And our additive is the minus 5 on the inside, which means we have a horizontal translation of 5, which would mean the graph would go 5 units to the right. Let's take a look at our next one. We have a vertical dilation by a factor of 5. That 1 third on the inside is going to give us a horizontal dilation by a factor of 3. And then we have a vertical translation of minus 8, moving the graph down 8. So you might be tested on these transformations from a graph, from the equation. Here we're going to look at what happens to a table of values if we have these translations.

So my first one here, we have a table of values for... f of x, and then we're going to do a table of values for g of x with the corresponding transformations. So this one, we just have a minus 5 on the inside, which means it's moving to the right 5 units. So my f of x values will not change. My input values are going to change.

So if we're moving all of these to the right 5, I'm going to add 5 to each of these numbers. So negative 2 plus 5 is 3. 0 plus 5 is 5, 4 plus 5 is 9, and 8 plus 5 is 13. Here, I'm going to the left 3 and down 2. So if I'm going left 3, I'm going to subtract 3 from all of my x's. Negative 5, negative 3, 1 and 5. Down 2, we're going to subtract 2 from all my y values.

So this will be at 3, 2, 1, and 0. Here I've got two things. This negative is going to be a reflection in the y-axis, sorry, x-axis, which will affect the y values. This 2 is a horizontal dilation.

by a factor of one half. That will affect the x values. So for the horizontal dilation of 1 half, that means these values are all going to get cut in half.

So this will be a negative 1, 0 will be 0, 4 is going to become 2, 8 is going to become 4. For the reflection in the x-axis, all of these are going to become opposite. Okay, next, rather than going from equation to table, we're going to go from table to equation. So we're going to kind of try to pick up on what's happening going from here to here. So looking at my x values, all of these are two bigger than this, so we're going right to. All my y values are opposite, so it's a reflection in the x-axis.

So my equation for g is going to be a negative f of x minus 2, since it's going to the right. Okay, here my y values are staying the same, but my x values are getting multiplied by 1 half. So my g of x is going to equal f of 2x to get that horizontal dilation.

Okay, next, transformations. Obviously, if we're... stretching and moving and translating graphs that can affect the domain and range. So we have a function that's domain goes from negative 8 to 4 and the range goes from negative 2 to 9 and we're going to find the domain and range of of the resulting graph. So here I've got g of x is f of x minus 2 plus 3, so we're going right 2 and up 3. So my domain will be affected by this, my range will be affected like this.

So if my domain is negative 8, 4, and I'm going to go to the right 2. We're going to add 2 to these. So my domain is going to be negative 6, 6. My range of negative 2, 9, if we're going up 3, we are going to add 3 to these. So negative 2 plus 3 is 1. 9 plus 3 is 12. Okay, here we have a reflection in the x-axis. We have a vertical dilation by a factor of 2. And then we're going to the left one and down three. Okay, so I'm just going to do this one step at a time.

We're going to look at what's happening to the domain versus what's happening to the range. So domain is going to be affected by the reflection, the vertical dilation, and the vertical shift. So reflection in the x-axis is going to take these... Sorry, these are going to affect my range.

So this is going to be a reflection, which is going to make this negative 2 become a positive 2, and this 9 to become a negative 9. But really, that just changes... my range to be going from a negative 9 up to a positive 2. My vertical dilation is going to multiply those values by 2. So now I'm going to be at negative 18 to 4. And then we're going to move it down 2, or down 3. So this will be at negative 21. and 1. The only thing that's affecting the domain is this left one, which means I'm going to subtract 1 from both of those. So that was a lot of questions on topic 112. That is a topic that you will see over and over and over again. So it's one you really want to get down.

Now we're going to look at 113 and 114, which is going to be all about modeling different situations. So my first situation is an open top box is to be made from a 30 by 18 rectangular piece of paper. or a piece of cardboard, by cutting equal squares of length x from each corner. So we're going to take a little chunk out of each corner.

And then fold those up, which is going to make our open-topped box. So we're told, write an equation for v of x, the volume, and what is the appropriate domain? Okay, well my volume is going to be length times width times height.

I'm going to start with the height because when I fold these guys up, my height is going to be x. My length is going to be this dimension here. Well, if this whole thing is 30 and I'm taking x out of here and x out of here, this is going to be 30 minus 2x.

And then similarly here, this is going to be 18 minus 2x. Because same thing, I'm taking an x chunk out here, an x chunk out here, so that's going to be 18 minus 2x. The domain is my set of x values.

So obviously I can't have an x value of like 100. I can't have an x value of negative 10. Those don't make sense in this problem. So at a minimum... My x has to be bigger than 0. Okay, at my maximum, and really that's going to be dictated by this side, the biggest chunk of x I could possibly take out would be 9 from both sides, but I'm just going to make it less than 9, because if it is 9, then we just kind of basically cut the whole thing in half. But if it's a little bit less than 9, we'll have a little bit of the box left, so that would be my domain.

Okay, next model we're going to make. An ice skating rink charges $10 for two hours of skate time. Each additional hour costs $3 to skate. Write a piecewise equation that gives the cost C to skate for T hours.

Okay, so this piecewise is going to have two equations. One for the first two hours and one for the additional hours. So we have a flat fee of $10. When we have been skating for 0 to 2 inclusive. If we stop at 2 hours, we're good with $10.

After that, we're going to get charged an additional $3. But we're only going to get charged the $3 for each hour after the initial 2. So for my equation, we're going to have... I have 3 for every t value bigger than 2. So for example, if I skated for 3 hours, I'd only want to add one $3 additional charge.

So that's where this comes in. This will be when t is greater than 2. We're not told any restrictions on the cap. Maybe you can just keep skating forever.

But with our information given, this would be our piecewise. For our last couple of questions, we will be using the calculator to do some models using the regression features of our calculator. I'm going to demonstrate this on the TI-84.

If you are using a different calculator, then you'll have to figure out how to do that with those calculators. And remember that Desmos is now going to be embedded during the AP exam, so if you learn how to do the regressions in Desmos, you can use that. Using the TI-84. So we're going to do a quadratic regression and use that to find or predict f of 20. So on the TI-84, you're going to start by pressing the start and we're going to go to edit. If you have data in here, you can highlight the list and do clear enter and that'll clear it.

Next, you want to populate your lists. Now that we've populated the list, we want to do the quadratic regression. So you're going to go... Okay, sorry.

I can read. Quadratic regression, we nailed it, stored it in y1. So if you go to y1, you're going to see the unrounded answer. I'm going to go ahead and just take these variables and plug them in.

I'm going to do three decimals, but we have it stored in our y equals. Okay, so taking those values and plugging them into the equation gives me that. That's not as accurate as we want it to be.

So when I evaluate for f of 20, we're going to go to y1, and we're going to evaluate at 20 because we stored our equation in 20. So if we use open parentheses, that will evaluate at the thing I want. So according to this, my f of 20 is approximately 751. Next we're going to do this a cubic regression for the second set of data. So again, stat, edit. We want to get rid of that so we're going to do clear, enter, clear, enter, and then enter our data.

If you ever get that error, dim mismatch, what it means is that your lists are not of the same length. So I accidentally had too many, so we're good. So good thing to know what those error codes mean. Now it's happy. So we copy that down, and then we want to evaluate this one for 13. So I'll do Y1, evaluate it at 13. So accurate to three decimals, we get 34.387.

So our last example, the number of zebras in a certain area are growing due to conservation efforts. The number of, so the table gives the number of zebras since 2018. The number of zebras can be modeled using a quadratic function. So we're going to write it in the form a t squared plus b t plus c.

So I've already entered my data. So I'll do a stack calc, quadratic regression again, store it, copy those values down. And lastly, we're going to use this to predict the number of zebras in the year 2024, which is equal to D equals 5. So we'll do Y1 because we stored it, evaluated it at 5. So we get 82.2, or I would round in the context would be 82 zebras.

So this was AP Precalculus Topics 112, 113, and 114. Thank you for watching.

Transcript for:AP Precalculus Unit 1 Review

Transcript for:
AP Precalculus Unit 1 Review