Exponential Functions. This information was covered in a previous section of the workbook, but it won t hurt to go over it again.


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1 Eponential Functions This inormation was covered in a previous section o the workbook, but it won t hurt to go over it again. Standard eponential unction = ca + k The c term is a constant that can make to graph relect about a horizontal ais or change to scale o the graph o the unction proportionately. I c is a positive value, then you will have a standard looking growth or decay curve. I c is negative, the growth or decay curve will lip upside down. We will get into the eects dierent values o h and k have on this unction shortly. What we will concentrate on here is identiying an eponential unction as being growth or decay, and inding the range, domain and key point o the unction. Eponential Growth Eponential Decay = ca + k = ca + k I a>1 I 0<a<1 Obviously i a= 1, we are raising 1 to various powers, and we wind up getting a horizontal line because no matter what you do, raising one to any power still yields a result o one. Pay special attention to the eponential decay unction. The statement 0< a< 1 is saying that the value o a is a raction whose value is between zero and one. Do not make the mistake o just looking or a raction to determine whether or not the unction is decay. Make sure the value o the raction is between zero and one. The values or variables,, h k and c act to make the graph shit let/right, up/down, change the scale or will relect the unction about a horizontal ais.
2 Notice the key point or each o these unctions is the point ( 0,1 ). This inormation is vital. This key point will shit depending on the values o h, k and c. To ind the value o the key point, evaluate h = 0. In other words, ind the value o that would create a problem such as 3 to the zero power. This number is the value o the key point. To ind the y value, substitute the value back in. Reer to the ollowing eample. 3 5 to ind the key point evaluate 3 = 0 3 = 0 = 3 this is the value o the key point now substitute 3 back into the problem or = 1+ 5 = 6 so the key point is ( 3,6) = ca + k As we work to translate these unctions, use ( 0,1 ), as the deault key point to any eponential growth or decay curve that is above the horizontal asymptote where the value o c is 1. In 0,1 to assist you in shiting the other words, i the value o c is positive one, use the point unction. I the graph o the unction is below the horizontal asymptote, and the c value is 0, 1 as the key point. I the value o c is any other number, you must 1, you will use ind the key point algebraically. Consider the eample above. 3 5 The irst thing I did was notice that the graph o this unction shits right 3 and up 5. Pay special attention to the value o k. That tells you where your new horizontal asymptote is going to be, in this case, at y= 5. Since the value o c is positive one, begin at the key point ( 0,1 ). Since the graph will shit right 3 and up 5, simply add 3 to the value, and 5 to the y value o the key point. This produces a new key point o ( 3,6 ). Observe how this inormation matches the work above. The key point in these unctions acts as the verte in a parabola. It gives you a point o reerence with which to shit the unction. Make sure the correct key point is used rom the beginning, either ( 0,1 ) or ( 0, 1) number. For eample, the unction = 3 has a key point o ( 0,3 ), not ( 0,1 ).. Remember, i the c term is a number other that 1 or 1, the key point is actually multiplied by that
3 Here we will look at how to graph these unctions by means o translation. Eponential Growth = ca + k = 3 4 = 3 + = 3 5 Notice that the horizontal asymptote is at y=. Since the graph o this unction is going to be above the ais, begin with the key point (0,1). This unction shits right 4 and up. The dots have been let on the graph so it would be easier to see. Adding 4 to the value o the key point, and to the y value, the new key point is at (4,3). Just remember where to begin, and do not cross the horizontal asymptote. In this unction, the value o k is 5. This tells you the new horizontal asymptote will be at y = 5. Since the value o the constant c is a positive one, begin with the key point (0,1). This unction will only shit down 5 spaces. Thereore, subtract 5 rom the y value o the key point which is 1. This results in: (1 5 = 4) thereore, the new key point is at (0,4). Graphing eponential unctions by translation is relatively simple. The most diicult part will be inding the and y intercepts as the intercept will involve the use o logarithms.
4 Eponential Growth = ca + k = 3 = Since the value o c in the equation o this unction is 1, we must begin with the key point o (0,1). This is the key point, because that value o c caused the graph to relect about the horizontal asymptote. The entire unction will shit up 4, so the new horizontal asymptote is y = 4. The curve is going to shit right and up 4. By adding to the value o the key point, and 4 to the y value, the new point can be ound at (,3). Notice the graph runs right though that point. Here we have something that looks like decay. The value o a in this unction is greater that one, so it should be growth. What really happened here, is the 3 laws o eponents went to work. Is the same thing as. The power o a power rule says this 3 1 can be seen as ( 3 ). This simpliies to 3 decay curve. OK, so we begin with a decay curve that has a key point o (0,1). Add to the value, and 1 to the y value o the key point, and the new key point is (,), with a horizontal asymptote o y = 1. 1, a
5 Eponential Decay = ca + k 1 = = 4 1 = + 3 This is an eponential decay curve. The original graph will lie above the ais. Thereore, begin with the key point (0,1). The key point will shit to the let 3, and down, so subtract 3 rom the value and rom the y value o the key point, and the coordinates o the new point will be at (3,1). The graph o this unction shits down, so the horizontal asymptote o this unction is y = . This is an eponential decay unction that is relected and lies below the horizontal asymptote. The initial key point here is (0,1). Since this graph will shit right 4 and up 3, add 4 to the value o the key point and 3 to the y value. This yields a result o (4,). Since the entire graph shited upwards 3 spaces, the horizontal asymptote is y = 3. Once again, when graphing, do not cross the horizontal asymptote. The translations o these unctions are very similar to that o other unctions we have seen. A point o reerence with which to shit is all that is needed. Most important is to make sure to always use the appropriate key point to start with. Draw the horizontal asymptote irst, that way the graph o the unction does not accidentally cross it.
6 = ca + k The domain o any eponential unction is (, ). The values o c and k terms will determine the range o the unction. Since the horizontal asymptote o an eponential unction is given by y=k, the value o k will determine where the horizontal asymptote o the unction lies, whereas the value o c will determine i the unction is above or below that asymptote. Be careul not to use brackets when describing the range o an eponential unction. The horizontal asymptote must not be touched, so only parenthesis may be used to describe the range in interval notation. Here is an eample o inding the and y intercept o an eponential unction. Finding the intercept. = Finding the y intercept. Begin by substituting 0 or Begin by substituting 0 or. + 0 = = 3 + log 4 = log 3 ( ) + log 4 log 3 log 4 = log 3 + log 3 log 4 log 3 = log 3 = = 3 4 = 9 4 = 5 Now divide both sides by log 3. log 4 log 3 log 3 = log 3 log 3 log 4 log 3 = log As you can see, this unction has an intercept o approimately (0.74,0), and a y intercept o (0,5).
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