4.3 The Graph of a Rational Function

Transcription

1 4.3 The Graph of a Rational Function Section 4.3 Notes Page EXAMPLE: Find the intercepts, asyptotes, and graph of + y =. 9 First we will find the -intercept by setting the top equal to zero: + = 0 so = -. We write (-, 0). To find the y-intercept, put in a zero for : 0 + y =. You will get 0 9 y = and we write 0,. 9 9 To find the vertical asyptote, set the botto equal to zero: 9 = 0 We get = ± 3. For the horizontal asyptote, we notice the highest power on the top is less than the highest power on the botto. Fro the previous section we know that the horizontal asyptote is autoatically y = 0. Now we are ready for the graph. First plot the intercepts. Then use dotted lines to draw in the asyptotes. The y-ais is one asyptote, so we have a horizontal dotted line at y = 0. The other asyptotes are = ± 3. These will be vertical lines going through 3 and -3 on the -ais. Now we need to graph. On the left and right sides of the graph we have two possible ways the graph can be drawn as the picture below shows. Notice that on the ends the graph will follow one asyptote, turn and follow the other asyptote. This is always the case with rational epressions: We need to chose whether the graph is above or below the -ais on each end. In order to do this we need to choose a test points. We need to choose a point that is less than negative 3 since we want to know what the graph is doing to the left of vertical asyptote, = -3. I will choose -4. We will put our test point into the original equation: y =. We get y =. Since this nuber is negative I know that the graph ust ( 4) 9 7 be below the -ais. Now let s test a point to the right of the vertical asyptote, = 3. I will choose 4 since this is greater than 3. Once 4 + again we will put 4 in for in the original equation: y = (4) 9 5 We get y = which is positive so this tells e the graph is above 7 the -ais. So now we know the graph looks like the following:

2 Section 4.3 Notes Page Now we need to take care of the iddle part of the graph. In between the two vertical asyptotes a rational graph will always look like one of the following: If we look at where our intercepts are that will help us choose one of the following. The only graph above that would fit the intercepts we are already plotted would be the third graph, so the following will be the our copleted graph: EXAMPLE: Find the intercepts, asyptotes, and graph of + 4 y =. + 5 ( + ) Let s factor this first: y =. We should always factor soething like this to see if anything ( 3)( + 5) cancels. Nothing does on this one so we will proceed to find the inforation. First we will find the -intercept by setting the top equal to zero: ( + ) = 0. We will get = 0 and = -. We will write our answer as (0, 0) and (-, 0). Since (0, 0) is an -intercept then autoatically this is our y-intercept as well. To find the vertical asyptote, set the botto equal to zero: ( 3)( + 5) = 0 We get = 3 and = -5. For the horizontal asyptote, notice that the highest power on top is the sae as the highest power on the an botto, so we know that y =. Here the a n = and b =, so the horizontal asyptote is y = =. b Now we are ready for the graph. First plot the intercepts. Then use dotted lines to draw in the asyptotes. There will be a horizontal line through y =. The other asyptotes are = 3 and = -5. These will be vertical lines going through 3 and 5 on the -ais.

3 Section 4.3 Notes Page 3 Now we are ready to draw the graph. Let s look at the part of the graph to the left of = -5. We need to decide whether the graph will be above or below the horizontal asyptote. We actually do not need to use test points for this. Notice that our only intercepts are between the two vertical asyptotes. This eans this is the only play the graph crosses the -ais. So when we look at the part of the graph where is less than -5 and since there are no intercepts there I know the graph has to be above the horizontal asyptote. The sae reasoning can be said about the part of the graph where is greater than 3. There are no -intercepts in the part of the graph where is greater than 3, so I know this part of the graph will also be above the horizontal asyptote. Now let s look between the two vertical asyptotes. Reeber there are only four types of graphs that could appear here: By the way the intercepts are located this tells e that only the first two graphs would work. How can we tell which one it is? We need to pick a test point. It can be any nuber between -5 and 3. I will test = -3. I will ( 3) + 4( 3) 7 put this in for in the original equation: y = =. Since this nuber is negative that ( 3) + ( 3) 5 30 eans when is -3 the graph should be below the -ais. This is only going to happen with the second graph above (upside down parabola). So now we can finish our graph:

4 EXAMPLE: Find the intercepts, asyptotes, and graph of y =. 4 Section 4.3 Notes Page 4 Let s factor this first: y =. Now we will find the -intercept by setting the top equal to zero: ( )( + ) = 0. We are done. We will write our answer as (0, 0). Again since (0, 0) is an -intercept then autoatically this is our y-intercept as well. To find the vertical asyptote, set the botto equal to zero: ( )( + ) = 0 We get = ±. For the horizontal asyptote, notice that the highest power on top is less than the highest power on the botto, so we know that the horizontal asyptote is autoatically y = 0. Now we are ready for the graph. We only have one point to plot this tie. Then we have our dotted lines to draw in the asyptotes. There will be a horizontal line through y = 0. The other asyptotes are = ±. These will be vertical lines going through - and on the -ais. Now we need draw the graph. Right now not uch inforation is given. We need to use test an value that is less than - and greater than - to see if the graph is above or below the -ais. First I will test = -3. We will put this in 3 3 for in the original equation: y = =. We get a ( 3) 4 5 negative nuber so I know the graph is below the -ais at this point. I also need to test a point greater than. I will 3 3 choose = 3. You will get y = =. So I know that the graph ust be above the -ais. What about the part of the graph between the two vertical asyptotes? We need to do test points here as well to deterine which of the four odels the graph resebles. I will test = and = -. We will not test those values: y = = This tells us that when = the graph is below the -ais. 4 3 Now if we test = - you will get y = =. So we know the graph is above the -ais at = -. ( ) 4 3 This tells us that the only graph that will work in the iddle is the third one.

5 EXAMPLE: Find the intercepts, asyptotes, and graph of 4 y =. Section 4.3 Notes Page 5 ( )( + ) Let s factor this first: y =. Now we will find the -intercept by setting the top equal to zero: ( )( + ) = 0. We get = ± ±,0.. well can write this as ( ) For the y-intercept you would put in a zero for. If you do then you will be dividing by zero, so there is no y- intercept in this case. To find the vertical asyptote, set the botto equal to zero: = 0. We only have one vertical asyptote here. For the horizontal asyptote, notice that the highest power on top is ore than the highest power on the botto, so we know that this is no horizontal asyptote. However we need to find the oblique asyptote by doing long division: We need to put in a 0 ter since it is issing. After we do the division we end up with an + 0. This is the equation of the line that the graph will follow. We need to set up 0 our graph by plotting the intercepts and the asyptotes. We will have one vertical 0 asyptote at = 0 and the oblique asyptote will be the line y =. 4 We see where the graph will cross the -ais. This tells us that the graph will be in these section. To the left of the vertical asyptote the graph will follow the line y = and turn turn towards (-, 0). Then it will follow the vertical asyptote. To the right of the vertical asyptote the graph will follow the vertical asyptote, turn through (, 0) and then follow the line y =.

6 EXAMPLE: Find the intercepts, asyptotes, and graph of y =. 6 Section 4.3 Notes Page 6 (3 + )( 4) Let s factor this first: y =. Notice that we can cancel out the 4 fro the top and botto. ( + 4)( 4) Whenever this happens you will have what is called a hole in the graph. Whatever part you can cancel you want to set this equal to zero. We will have 4 = 0. Solving it we get = 4. This is not a vertical asyptote. There will be a hole in the graph at the value of So the equation we will now look at to get the inforation will be y =. + 4 To find the -intercept by setting the top equal to zero: 3 + = 0. We get =, so we have, (0) + For the y-intercept you would put in a zero for. If you do then you will have y = =. So 0,. (0) + 4 To find the vertical asyptote, set the botto equal to zero: + 4 = 0 and we get = -4. We only have one vertical asyptote here since the other ter canceled out. We already deterined that there will be a hole at the value of 4. To find the horizontal asyptote we see the highest power on top is the sae as the highest power on the an 3 botto. So the horizontal asyptote is y =. Here the a n = 3 and b =, so y = = 3. b So we have drawn in the asyptotes and the intercepts. We notice that there are no -intercepts to the right of = -4. This tells us that the graph ust be above the horizontal asyptote (y = 3). If the graph was below this line then it would have crossed the -ais but we know this does not happen. To the right of the = -4 we notice the graph does cross the -ais so we know the graph is in this lower portion and not above y = 3. Of course if you are unsure where the graph should be you can always use tests points. You test a point less than = -4 and greater than = -4. When we draw the graph we need to ake sure that we put a hole when = 4.