WEBVTT 00:22.000 --> 00:30.064 In this section, we're going to learn how to graph piecewise-defined functions, and we're also going to discuss 00:30.097 --> 00:39.073 shifting and reflecting graphs of functions. Let's begin with piecewise-defined functions. A piecewise-defined 00:39.106 --> 00:49.316 function is a function defined by two or more expressions. This is an example of a piecewise function. 00:49.360 --> 01:04.064 It says that f of x is defined as three when x is less than negative two and f of x is defined as negative one when x 01:04.098 --> 01:13.908 is greater than or equal to negative two. Let's graph this piecewise-defined function on our coordinate plane. 01:13.941 --> 01:24.018 Let's think for a moment about the ordered pairs that each of these pieces of this function represent. 01:24.080 --> 01:35.162 If I choose an x-value, for instance, let's say negative five. If x is negative five, then the piece of this function that I 01:35.196 --> 01:43.604 must use to define the function is f of x equals three because this is the piece of the function 01:43.637 --> 01:54.515 that is defined when the x-value is less than negative two. Thus, the y-value, or the function value here, is three. 01:54.560 --> 02:05.926 If I choose another x-value that is less than negative two, the function value, or the y-value, will also again be three. 02:05.960 --> 02:22.743 So negative four, three. Negative three, three. These are ordered pairs that are on this function that are represented 02:22.776 --> 02:33.854 by this top piece. Now once x gets to negative two, this top piece is no longer the definition of the function 02:33.920 --> 02:41.395 because this top piece is only defined as the function value when x is less than negative two. 02:41.440 --> 02:51.005 When x is greater than or equal to negative two, the function is defined as negative one. 02:51.040 --> 03:04.919 So when x is negative two, the y-value, or the function value, is negative one. 03:04.960 --> 03:12.126 And for any x-value that is greater than negative two, the function is so defined as 03:12.160 --> 03:18.999 f of x equals negative one. So I can choose any other x-value greater than negative two, 03:19.040 --> 03:35.382 and the y-value will be negative one. Let's plot these points. Here we have negative five, three. 03:35.440 --> 03:49.096 We have negative four, three. We have negative three, three. We can see that these ordered pairs are lying on a horizontal line. 03:49.129 --> 03:57.004 We should also recognize the equation of a horizontal line is of the form y equals a number. In this case, 03:57.040 --> 04:04.678 f of x is equal to three represents a horizontal line with the y-values of three. So this top piece 04:04.720 --> 04:11.952 is a horizontal line for all x's that are less than negative two. So for all of these 04:12.000 --> 04:23.163 x values that are less than negative two, the y-value is three. But when I get to negative two, 04:23.200 --> 04:34.441 I'm going to put an open circle there because the function value is not defined as three when x is negative two. 04:34.480 --> 04:41.381 Once x becomes negative two, the function now is defined as the y-value equal to negative one. 04:41.440 --> 04:48.922 And again, this is a horizontal line y equals a number. All of the function values are negative one 04:48.960 --> 04:55.120 when the x's are greater than or equal to negative two. So I can see these ordered pairs and 04:55.129 --> 05:02.102 an infinite number more ordered pairs all who have the same y-value of negative one, 05:02.160 --> 05:10.677 beginning with x equals negative two. We'll put a closed dot at negative two to show that when 05:10.720 --> 05:20.654 x is equal to negative two, this is the value of the function. And for all of these other x-values, 05:20.720 --> 05:33.801 the y-value is negative one. This is the graph of the piecewise function defined here. 05:33.840 --> 05:42.342 Let's take a look at another piecewise-defined function. This function is defined as three x plus 05:42.400 --> 05:51.952 one, when x is less than or equal to zero, and it is defined as five minus x when x is greater than 05:52.000 --> 06:00.494 zero. So let's begin by thinking about this first definition of this piecewise-defined function. 06:00.560 --> 06:07.067 You may recognize it in the form of m x plus b. So this is a linear function or a graph 06:07.120 --> 06:15.876 of a linear function for this first piece. It has a slope of three and a y-intercept of one. 06:15.920 --> 06:23.951 We're only going to graph this line for x's that are less than or equal to zero. 06:24.000 --> 06:30.757 The y-intercept of one is where we'll start, which is at an x-value of zero. And because zero 06:30.800 --> 06:40.434 is included in this function, less than or equal to, we put a closed dot at the y-intercept of one. 06:40.480 --> 06:47.241 The slope here is three. Now recall when we graph a line, and we have a slope, we can 06:47.280 --> 06:56.583 go up three for this and to the right one. Or we can go down three and to the left one 06:56.640 --> 07:04.658 because negative three divided by negative one is equivalent to three over one, which is 07:04.720 --> 07:12.499 our rise over run. Because I want to graph the values where x is less than or equal to zero, 07:12.560 --> 07:21.875 we'll go down three and left one from our y-intercept, giving us the second point here. 07:21.920 --> 07:27.614 I can graph a third point to make drawing the line a little simpler, 07:27.680 --> 07:41.295 and this would be the third point on our graph. I'll connect these points with a straight line. 07:41.360 --> 07:48.835 And now we'll discuss this bottom portion of our piecewise-defined function. 07:48.880 --> 07:55.542 This is telling us that the function is defined as five minus x when x's are greater than zero. 07:55.600 --> 08:06.019 Again, recognize this is in the form or can be in put in the form m x plus b, 08:06.080 --> 08:18.098 where m would be negative one and b would be five. So we have a y-intercept of five 08:18.160 --> 08:27.007 and a slope of negative one for this bottom piece of our piecewise-defined function. 08:27.040 --> 08:35.549 My y-intercept is five, but recall that this function is only valid for x is greater than zero. 08:35.600 --> 08:49.796 So at zero we will have an open circle as opposed to a closed circle at five. Our slope again here is negative one, 08:49.840 --> 08:57.471 so we can have a rise of negative one, meaning down one, and to the right one. 08:57.520 --> 09:04.444 Or we can have a rise of positive one and to the left one. These are equivalent. 09:04.480 --> 09:12.419 Because I want to graph to the right of this open circle, I'm going to choose to go down one and 09:12.480 --> 09:20.127 to the right one because to the right of x equals zero are the values where x is greater than zero. 09:20.160 --> 09:27.801 Down one and to the right one. Down one and to the right one. And now I have three points, 09:27.840 --> 09:45.886 making it easy to draw the line that goes through these points, and this is the graph of the piecewise-defined function. 09:45.920 --> 09:51.391 Now let's consider the function f of x equal to the absolute value of x. 09:51.440 --> 09:58.999 We are going to complete the table and then use the data in the table to graph the function. 09:59.040 --> 10:07.140 Let's begin with the x-value negative two. When x is negative two, f of x is equal to 10:07.200 --> 10:14.915 the absolute value of negative two, and the absolute value of negative two is positive two. 10:14.960 --> 10:25.392 When x is negative one, f of x is the absolute value of negative one, which is positive one. 10:25.440 --> 10:37.704 When x is zero, f of x is the absolute value of zero, which is zero. When x is one, f of x is the 10:37.760 --> 10:46.513 absolute value of one, which is one. And when x is two, f of x is the absolute value of two, 10:46.560 --> 10:59.726 which is two. Let's plot these ordered pairs on our graph. We'll begin with negative two, two. 10:59.760 --> 11:12.672 Next, we have negative one, one. Zero, zero which is here at the origin. 11:12.720 --> 11:24.751 We have one, one, and two, two. Now as we think about the shape of this graph, 11:24.800 --> 11:31.458 let's think about what would happen if we extended this table in both directions. For the x- 11:31.520 --> 11:38.431 values that continue greater than two, these are all positive values, and the absolute value 11:38.480 --> 11:47.140 of a positive value is a positive value. In this case, the x- and the y-values will be the same. 11:47.200 --> 11:56.182 So this right-hand side of the graph will continue in a linear format with a slope of positive one, 11:56.240 --> 12:04.190 where we go up one and over one. All of the x-values here have the same y-value 12:04.240 --> 12:14.267 for their function value. And so the graph of this side of the absolute value of x 12:14.320 --> 12:27.814 looks like this. It looks like a linear function with a slope of one. Now on this side where we have negative x-values, 12:27.847 --> 12:35.955 notice that the function value has the opposite sign of the x-values. So where the x-values are negative, 12:35.989 --> 12:44.431 but when I take the absolute value of those negative numbers, I get a positive value. And so these y-values 12:44.464 --> 12:54.908 are the opposite sign of the x-values, and they continue in that same pattern on the left side of zero. 12:54.960 --> 13:06.052 So as I graph this side of the absolute value of x, I can see that this side, left of zero, 13:06.086 --> 13:13.793 looks like a linear function that has a slope of negative one, down one and to the right one for 13:13.840 --> 13:23.036 each of these ordered pairs. The overall shape of this absolute value function is the shape of a v. 13:23.069 --> 13:31.644 However, we can define this absolute value function as a piecewise-defined function. 13:31.680 --> 13:36.816 Since we notice that the right side looks like a linear function with a slope of 13:36.880 --> 13:44.524 one and the left side looks like a linear function with a slope of negative one, 13:44.560 --> 13:52.365 then we can define f of x which is the absolute value of x as a 13:52.400 --> 14:04.277 piecewise-defined function where f of x is equal to x when x is greater than zero 14:04.320 --> 14:16.823 and f of x is equal to negative x when x is less than or equal to zero. 14:16.880 --> 14:22.362 Both of these definitions the absolute value of x and this piecewise-defined 14:22.400 --> 14:33.640 function represent the same function and the same graph, the absolute value of x. 14:33.680 --> 14:40.914 Consider the two functions f of x equals the absolute value of x and g of x equals the absolute 14:40.960 --> 14:47.854 value of x plus three. Complete the table, and then graph the functions on the same axes. 14:47.920 --> 14:54.260 Then we'll describe the relationship between the graph of f and the graph of g. Well, we just 14:54.320 --> 15:01.534 recently graphed f of x equals the absolute value of x. So here are those function values completed. 15:01.600 --> 15:08.441 And here's the graph of f of x equals the absolute value of x. Recall that that was the shape of a v 15:08.480 --> 15:15.481 that could be also written as a piecewise-defined function. Now let's do the table for g of x. 15:15.520 --> 15:22.422 If x is negative two, then g of negative two is the absolute value of negative two, 15:22.480 --> 15:34.734 which is two, plus three, giving us five. If x is negative one, g of negative one is the absolute value of negative one, 15:34.767 --> 15:51.084 which is one, plus three, or four. When x is zero, g of x is the absolute value of zero, which is zero, plus three. 15:51.120 --> 16:04.497 When x is one, g of one is the absolute value of one, which is one, plus three. And if x is two, g of two is the 16:04.530 --> 16:21.714 absolute value of two, which is two, plus three which is five. Let's plot these ordered pairs. Negative two, five. 16:21.760 --> 16:51.778 Negative one, four. Zero, three. One, four. And two, five. We notice the same type of pattern showing up 16:51.811 --> 17:02.922 on this red graph of g of x as we see on the green graph f of x. In other words, the right side appears to be 17:02.955 --> 17:25.144 linear with a slope of positive one. And the left side also appears to be linear. It has a slope of negative one. 17:25.200 --> 17:32.819 These two graphs, f of x and the graph of g of x, have the same shape. They both still 17:32.880 --> 17:40.026 have the shape of the v, the absolute value of x shape. We're going to describe the relationship 17:40.080 --> 17:49.402 a little more precisely now. While they have the same shape, notice that the red graph is higher 17:49.440 --> 18:03.082 or vertically shifted from the green graph. In fact, every ordered pair is one, two, three units 18:03.120 --> 18:12.658 higher than the corresponding ordered pair of the green graph. Every ordered pair is three 18:12.720 --> 18:35.948 units higher. So we say that the red graph g of x has been vertically shifted three units. 18:36.000 --> 18:56.903 This was the graph of g, or g of x, is a vertical shift of three units from the graph of f. 18:56.960 --> 19:07.647 As we noticed in the last problem, g of x was a vertical shift three units up from f of x, 19:07.680 --> 19:15.454 and the difference in the equations of f of x and g of x was this plus three. 19:15.520 --> 19:21.861 Let's look at our rules for vertical shifts. Let c be a positive number. 19:21.920 --> 19:33.940 The graph of g of x equals f of x plus c is the graph of y equals f of x shifted c units upward. 19:34.000 --> 19:39.478 This makes sense based on what we saw here. Our c value was three. 19:39.520 --> 19:47.320 Therefore, since we had a plus three, this graph shifted up three units. 19:47.360 --> 19:58.364 The graph of g of x equals f of x minus c is the graph of y equals f of x shifted c units downward. 19:58.400 --> 20:03.235 So if this problem had been the absolute value of x minus three, 20:03.280 --> 20:10.910 we would expect to see the absolute value of x graph shifted down three units. 20:10.960 --> 20:21.220 Let's do one more example. Let's graph h of x equals the absolute value of x minus five. Here 20:21.280 --> 20:28.027 we know that f of x is the absolute value of x. That is our green graph. And here we 20:28.080 --> 20:40.039 are subtracting five from f of x, so h of x can be defined as f of x minus 20:40.080 --> 20:49.949 five. Meaning that all of the ordered pairs on f of x are going to be shifted downward five units. 20:50.000 --> 20:57.723 I can choose any of the ordered pairs to make this shift. I'll start here at negative two, 20:57.760 --> 21:09.535 two. If I shift this down five, this puts me at the ordered pair negative two, negative three. 21:09.600 --> 21:19.879 I can shift this ordered pair, which is at the origin, down five, and I will be at negative five 21:19.920 --> 21:27.219 on the y-axis. And if I shift this ordered pair, which was at two, two, 21:27.280 --> 21:37.062 down five units, then I will be at two, negative three. And it's going to maintain the same shape 21:37.120 --> 21:47.406 as the graph of f of x. So it will still have the v shape. So on the right side, we will expect this 21:47.440 --> 21:59.819 linear shape with a slope of one. And on the left side, we will have this linear shape on the graph with a 21:59.852 --> 22:09.061 slope of negative one. Every ordered pair from the green graph has been shifted down five units 22:09.120 --> 22:17.937 to obtain the graph of h of x which is the absolute value of x minus five. 22:18.000 --> 22:24.577 Consider the functions f of x equal the absolute value of x and g of x equals the absolute value 22:24.640 --> 22:33.219 of x minus two. We're going to complete the table and then graph f and g on the same axes. 22:33.280 --> 22:39.391 And we will describe the relationship between the graph of f and the graph of g. We've 22:39.440 --> 22:44.663 already graphed the values for f of x equal to the absolute value of x. And here is the 22:44.720 --> 22:50.836 graph that we've seen in previous examples. Now let's complete the table for g of x. 22:50.880 --> 22:59.311 When x is negative two, g of negative two will be the absolute value of negative two minus two, 22:59.360 --> 23:07.553 which is the absolute value of negative four. The absolute value of negative four is positive four. 23:07.600 --> 23:13.158 g of negative one is the absolute value of negative one minus two, 23:13.200 --> 23:23.002 or the absolute value of negative three which is equal to three. g of zero is the absolute value of 23:23.035 --> 23:34.446 zero minus two, which is the absolute value of negative two. The absolute value of negative two is two. 23:34.480 --> 23:41.487 g of one is the absolute value of one minus two, which is the absolute value of negative one, 23:41.520 --> 23:55.801 which is one. g of two is the absolute value of two minus two, which is zero, and the absolute value of zero is zero. 23:55.840 --> 24:03.075 g of three is the absolute value of three minus two, which is the absolute value of one, 24:03.120 --> 24:10.683 which is one. And finally, g of four is the absolute value of four minus two, 24:10.720 --> 24:19.058 which is the absolute value of two, which is equal to two. Now let's plot these points on our graph. 24:19.091 --> 24:55.227 We have negative two, four. Negative one, three. Zero, two. One, one. Two, zero. Three, one. And four, two. 24:55.280 --> 25:04.603 Connecting these three points with a line of slope one gives us the right side of this absolute value 25:04.640 --> 25:13.312 graph. And connecting this left side with a straight line and a slope of negative one 25:13.360 --> 25:24.590 gives us the left side of this graph. So here you can see the graph of f and the graph of g. 25:24.640 --> 25:31.864 Now let's see how they compare to each other. They have the same shape, but all of the ordered pairs 25:31.920 --> 25:43.308 on f have been shifted to the right. By how many units? two. Notice that our function was g of 25:43.360 --> 25:52.017 x equals absolute value of x minus two, and every ordered pair here has been shifted to the right 25:52.080 --> 26:02.895 two units from the green graph, which was f, to the red graph, which is g. So we say that g 26:02.960 --> 26:27.619 is a horizontal shift of two units to the right when compared to graph f. 26:27.680 --> 26:33.325 Reflecting on our last examples of horizontal shifts, we can now look at the rules for 26:33.360 --> 26:42.267 horizontal shifting. Let c be a positive number, the graph of g of x equal to f of x plus c 26:42.320 --> 26:52.511 is the graph of y equals f of x shifted c units to the left. Now this may seem a bit counterintuitive. 26:52.560 --> 26:59.318 Here we are adding a positive number to x, but yet we are shifting 26:59.360 --> 27:07.559 to the left. We can think about why that may be. We're trying to get the same output value 27:07.600 --> 27:18.904 of f of x. Well, we need a value that is c units less than the given input of f of x. 27:18.960 --> 27:29.381 The graph of g of x equal to f of x minus c is the graph of y equals f of x shifted c units to the 27:29.440 --> 27:38.824 right. Again notice, we have x minus a positive number, and we're shifting to the right. Again, 27:38.880 --> 27:46.198 we're looking for the same output value with a minus sign inside of our grouping symbols here, 27:46.240 --> 27:55.073 for x minus a positive number. And so in order to get that, we have to shift this graph to the right. 27:55.120 --> 28:01.480 Let's take a look at these two examples. We're going to start by graphing the function h of x 28:01.520 --> 28:11.156 equals the absolute value of x plus four. Which rule applies here? Well, we're adding a positive 28:11.200 --> 28:18.430 number to x inside of our grouping symbols, or the absolute value symbols. This represents 28:18.480 --> 28:28.840 f of x plus c. Therefore, we're going to shift c units to the left, where c is four. 28:28.880 --> 28:35.247 Here's our function f of x equal to the absolute value of x, and we're going to shift it c units or 28:35.280 --> 28:44.423 four units to the left. So starting with our vertex we will count four units to the left. 28:44.480 --> 28:49.294 And again, the right side of this absolute value function is going to look the same 28:49.360 --> 28:57.636 it's going to have a linear shape with a slope of one. And you can see from this graph 28:57.680 --> 29:04.342 that, or this portion of the graph, that these points are all shifted one, two, three, 29:04.400 --> 29:12.751 four units to the left. Let's complete the other side of the graph now. The left side of the graph 29:12.800 --> 29:23.962 has this linear shape with a slope of negative one. All of the points on this graph of f of x have been shifted to the left 29:24.000 --> 29:34.840 four units. This is the graph of h. Let's now take a look at k of x. k of x 29:34.880 --> 29:44.816 is the absolute value of x minus five plus three. So we have to apply two of our rules here. First, 29:44.880 --> 29:51.323 inside of our absolute value symbols, we see x minus a positive number, so 29:51.360 --> 29:59.064 our second rule here will apply for shifting. This is the graph of f of x minus c. So this 29:59.120 --> 30:07.405 minus five inside of our absolute value symbols will shift the graph to the right five units. 30:07.440 --> 30:13.645 Recall that when we add a number outside of the grouping symbols, or in this case 30:13.680 --> 30:21.086 the absolute value symbols, then we shift either up or down. When we have a positive value c, 30:21.120 --> 30:26.925 we shift up. So in addition to the absolute value of x graph being shifted five units 30:26.960 --> 30:37.235 to the right, we're also going to shift this graph three units up. Let's start with our absolute value function, 30:37.269 --> 30:54.920 and we will shift five units to the right and three units up. And now we'll complete both sides of our graph. To the right 30:54.960 --> 31:15.307 and then to the left. Every point on the original graph of f is now shifted five units to the right 31:15.360 --> 31:30.922 and three units up. So this is the graph of k. Consider the functions f of x equals the absolute value of x and g of x 31:30.956 --> 31:39.764 equals the negative absolute value of x. We're going to complete the table and graph f and g on the same axis. 31:39.798 --> 31:47.138 And then we're going to describe the relationship that we notice between the graph of f and the graph of g. 31:47.200 --> 31:52.844 The graph of f of x equals the absolute value of x is already graphed for us here, 31:52.880 --> 32:00.685 and the table is completed. Let's focus on g of x. First, when x is negative two, 32:00.720 --> 32:08.660 we're looking for g of negative two. This is the negative absolute value of negative two. 32:08.720 --> 32:13.298 The absolute value of negative two is positive two, and then we're multiplying 32:13.360 --> 32:24.109 that value by negative one, giving us a value of negative two. g of negative one is the negative 32:24.160 --> 32:31.216 absolute value of negative one. The absolute value of negative one is positive one. 32:31.280 --> 32:40.658 Then multiplied by negative one gives us g of negative one equal to negative one. 32:40.720 --> 32:51.569 g of zero is the negative absolute value of zero. The absolute value of zero is zero, and zero times anything is still zero. So 32:51.603 --> 33:05.483 g of zero is zero. g of one is the negative absolute value of one. That is negative one. 33:05.520 --> 33:17.228 And g of two is the negative absolute value of two, which is negative two. And now let's plot these ordered pairs on our 33:17.262 --> 33:32.911 graph. We'll begin with negative two, negative two. Next, we have negative one, negative one. Zero, zero. 33:32.960 --> 33:47.692 One, negative one. And two, negative two. Let's connect these points to form our graph. On the right side of our graph, 33:47.725 --> 34:00.105 we have this linear shape with a slope of negative one. And on the left side of our graph, we also have a linear 34:00.160 --> 34:15.220 shape, but this side has a slope of positive one. Now let's compare these two graphs of f and g. 34:15.280 --> 34:22.594 All of the ordered pairs on f have positive y-values, while all of the ordered pairs on g 34:22.640 --> 34:36.207 have negative y-values. This appears to be a mirror reflection of f across the x-axis. So we say that g is a 34:36.241 --> 35:09.641 reflection of f about the x-axis. Reflections about the x-axis. The graph of g of x equal to negative f of x is the graph of y 35:09.680 --> 35:17.782 equals f of x reflected about the x-axis. Let's look at this next example. We are to graph 35:17.840 --> 35:25.557 h of x equal to the negative absolute value of x plus four minus two. 35:25.600 --> 35:33.298 Let's look at the different changes to the graph of f of x that we see here. In h of x, 35:33.360 --> 35:38.036 I notice that there is a negative outside of the grouping symbols or outside of the absolute 35:38.080 --> 35:45.877 value symbols. There is a plus four inside of the absolute value symbols. And there is a minus two 35:45.920 --> 35:53.551 outside of the absolute value symbols. If you were to evaluate this function for a given value of x, 35:53.600 --> 36:02.126 where would you begin? You would begin by adding four to the value of x. That's 36:02.160 --> 36:08.266 what we'll begin with this transformation. The plus four inside of the absolute value symbol 36:08.320 --> 36:15.873 means that we're going to shift the graph of the absolute value of x to the left four units. 36:15.920 --> 36:22.447 The next thing we would do if we were evaluating this function is to multiply that value by negative one. 36:22.480 --> 36:31.489 This negative in front of our absolute value symbols is a reflection of the graph of f of x about the x-axis. 36:31.522 --> 36:39.630 And the last thing that we would do if we were evaluating this function for a given value of x is subtract two. 36:39.664 --> 36:49.474 And this is a vertical shift of negative two. So we're going to shift the graph of f of x down two. So we have 36:49.520 --> 36:59.217 a horizontal shift left four, a reflection about the x-axis, and a vertical shift of negative two. 36:59.280 --> 37:07.225 Let's put that on our graph. Beginning here with our vertex, we'll shift left four. 37:07.280 --> 37:14.265 A reflection about the x-axis does not change that because zero is neither positive nor negative. 37:14.320 --> 37:24.709 And we have minus two of a vertical shift. I could take any other ordered pair and make the same transformation, 37:24.742 --> 37:31.149 or knowing the shape of this graph, I could complete the graph at this point. Let's choose another ordered 37:31.200 --> 37:40.758 pair. Here's a point on our parent graph one, one. I would shift this point to the left four. 37:40.800 --> 37:46.030 It has a reflection about the x-axis, meaning that the y-value would now be 37:46.080 --> 37:52.270 negative one instead of positive one. And I have a vertical shift of negative two, 37:52.320 --> 37:59.010 which will shift us down two units from there. And now you can clearly see 37:59.043 --> 38:08.753 that this side of the graph will have this linear shape with a slope of negative one. 38:08.800 --> 38:18.262 And the other side of the graph will also have a linear shape. It will have a slope of positive one. 38:18.320 --> 38:25.536 And this is the graph of h, where every ordered pair from the graph of f has the same 38:25.600 --> 38:45.840 transformation shifted left four, reflected about the x-axis, and shifted down two units.