Power functions: f(x) = x n, n is a natural number The graphs of some power functions are given below. n even n odd


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1 5.1 Polynomial Functions A polynomial unctions is a unction o the orm = a n n + a n1 n a 1 + a 0 Eample: = The domain o a polynomial unction is the set o all real numbers. The intercepts are the solutions o the equation a n n + a n1 n a 1 + a 0 = 0 The yintercept is y = 0 = a 0. The graph o a polynomial unction does not have holes or gaps we say that a polynomial unction is continuous and it does not have sharp corners or cusps we say it is smooth Power unctions: = n, n is a natural number The graphs o some power unctions are given below n even n odd Notice that when n is even, a power unction y = n behaves like a parabola graph is symmetric about the yais and contains points 1,1, 0,0, 1,1. When n is odd, a power unction y = n, n > 1 has the graph similar to the cube unction symmetric about the origin, contains the points 1,1, 0,0, 1,1. Power unction = a n, a 0 The graph o = a n is obtained rom the graph o y = n by stretching by a actor o a, i a is positive, and stretching by the actor o a and relecting about the ais, i a is negative. n even n odd
2 Zeros o a polynomial unction I r is such a number that r = 0, then r is called a zero o the unction. I r is a zero o a polynomial unction, then we have the ollowing: i r = 0 ii r, 0 is an intercept o iii r is a actor o, that is = rq q is the quotient in the division r We say that r is a zero o multiplicity n, i n is the largest power, such that = r n q Eample: Let = Note that 1 = = = 0. Thereore, r = 1 is a zero o. This also means, that 1,0 is an intercept and that 1 is a actor o, that is, when is divided by 1 the remainder is 0. To ind the other actor, q, we perorm the division Thereore, = = What is the multiplicity o that zero? Since 1 is a actor o, then the multiplicity o r = 1 is at least 1. I 1 had the multiplicity two, then would have 1 as a actor. Which means that the quotient q above, q = , would have 1 as a actor. I 1 were a actor o q, then 1 would be a zero o q. But q1 = = 3 0. This means that 1 is not a actor o q and consequently, 1 is not a actor o. Hence the multiplicity o 1 is one. Eample: Find all zeros o unction = and determine their multiplicity To ind zeros, solve the equation = = 03 = 0 or + = 0 or +3= 0 or  5 = 0 So the zeros are : 3, , 3, 5 To ind the multiplicities: i actor the polynomial completely; use eponents to indicate multiple actors ii The multiplicity o a zero r is the eponent o the actor r that appears in the product = zeros multiplicity 1 1
3 Suppose r is the zero o a polynomial unction. Remember that r,0 is the intercepts then I r is a zero o even multiplicity, then the graph o will touch the ais at the intercept r, 0 as shown below. or I r is a zero o odd multiplicity, then the graph o will cross the ais at the intercept r,0 as shown below. Multiplicity one Multiplicity larger than 1 odd or or One o the theorems o algebra says that every polynomial can be actored in such a way that the only actors are : a a number the leading coeicient b r n, where r is a zero with multiplicity n c +b +c m, where + b + c is prime
4 Remark: This theorem says that such actorization is possible, but it does not say how to obtain such actorization. End behavior o a polynomial unction When is large is large positive or large negative, then the graph o = a n n + a n1 n a 1 + a 0 behaves like the graph o y = a n n, where a n is the leading coeicient and n is the degree o. n odd a n positive a n negative n even a n positive a n negative Eample: Determine the degree and the leading coeicient o the polynomial unction = Give the equation o the power unction that the unction behaves like or with large absolute value. = This polynomial is already actored. The leading coeicient is 3. The degree can be obtained by adding the highest eponents o rom each actor actor degree Degree = = 9 Thereore, or large, behaves like y = 3 9.
5 Sketching the graph o a polynomial unction I use transormations, when possible II I the transormations cannot be perormed, use the inormation above to sketch the graph Eample: Graph = We can graph this unction using transormations The order o transormations is as ollows 1 graph basic unction y = 5 Shit the graph one unit to the right to obtain y = Stretch the graph by a actor o, to obtain y = 1 5 Shit the graph up by 3 units to obtain y = Eample: Graph = Transormations cannot be used i Determine the zeros, i any, o, their multiplicity and the behavior o graph near each zero = zeros 1  multiplicity 3 behavior o graph touches ais at 1 ii Determine the end behavior o Crosses ais at  like a cubic unction We need the leading coeicient and the degree o = Leading coeicient =
6 Degree: actor degree 0 3 Degree o = 0++3 = 5 behaves as y = 5 or large iii Use the inormation rom i and ii to draw the graph. The graph will start in the third quadrant green piece. It will continue to the irst zero  at which it will cross the ais like a cubic unction red piece. The graph will increase or a while, but then it will have to turn to reach the second zero 1, at which it will touch the ais red piece. Since there are no more zeros, the graph will continue, eventually to reach green piece that depicts its behavior or the large positive. Eample: Graph = +1+ Note that + 1 is always positive, so can be zero only when = 0 or =  zeros 0  multiplicity behavior o graph touches ais at 1 Touches ais at  End behavior: Leading coeicient = 1 actor degree
7 Degree o = + + = 6 and behaves like y = 6 or large, that is, it looks as below The graph o the unction is below Remarks:  We don t have enough inormation to know at what points eactly the graph will change the direction these points are called turning points. You can learn this in Calculus. But, we know that there are at most n1 turning points n is the degree o the polynomial  Though we know how the graph o a polynomial unction behaves or large positive and negative values and close to its zeros, we need Calculus to determine its behavior in between. In this course however, we ll assume that nothing etraordinary takes place.
8 5. Properties o Rational unctions A rational unction is a unction o the orm n n1 polynomial p an an 1 a1 a0 k k1 polynomial q bk bk 1 b1 b0 Eample The domain o a rational unction is the set o all real numbers ecept those, or which q = 0 To ind the domain: i solve q = 0 ii Write D = { q 0} 3 5 Eample: Find the domain o 1 i Solve: denominator = ii D = { 5 } =, 5 5, 5 5, A rational unction oten has asymptotes: vertical and/or horizontal/oblique. Inormally speaking, an asymptote is a straight line vertical, horizontal or slanted toward which the graph comes near. 5 Vertical and horizontal asymptotes vertical and oblique asymptotes vertical and horizontal asymptotes How to ind asymptotes Vertical: 1. Reduce to the lowest terms: i actor completely the numerator and the denominator; ii cancel common actors. Solve the equation: denominator = 0 3. I = r is a solution ound in, then the line = r is a vertical asymptote Horizontal: a i the degree o the numerator < the degree o the denominator, then the line y = 0 is the horizontal asymptote
9 a b i the degree o the numerator = the degree o the denominator, then the line y = b the horizontal asymptote n k is c i the degree o the numerator > the degree o the denominator, then the graph does not have a horizontal asymptote, however, i Oblique: d the degree o the numerator = 1 + the degree o the denominator, then the line y = quotient obtained by dividing the numerator by the denominator is an obliqueslanted asymptote. Remarks: 1. A rational unction can have only one horizontal/oblique asymptote, but many vertical asymptotes.. I a rational unction has a horizontal asymptote, then it does not have an oblique one. 3. The graph o a rational unction can cross a horizontal/oblique asymptote, but does not cross a vertical asymptote. Horizontal/oblique asymptotes describe the behavior o unction or with large absolute value; vertical asymptotes describe the behavior o unction near a point Eample: Find the asymptotes or the ollowing unctions 3 5 a 6 Vertical asymptote: 1 is in lowest terms 6 = 0 = 6 = 3 3 vertical asymptote: = 3 Horizontal/oblique asymptote: 3 degree o numerator 1 = degree o the denominator1, y is the horizontal asymptote 5 1 b Vertical asymptote: is in lowest terms numerator can t be actored = = 0 = 0 or  = 0 = 0 or = 3 vertical asymptotes : = 0, = Horizontal/oblique asymptote:
10 degree o numerator < degree o the denominator3, y = 0 is the horizontal asymptote c Vertical asymptote: 1 is in lowest terms the denominator cannot be actored + = 0 =  not possible no solution 3 vertical asymptotes : none Horizontal/oblique asymptote: degree o numerator 5 > degree o the denominator, there is no horizontal asymptote degree o numerator degree o the denominator, there is no oblique asymptote d Vertical asymptote: 1 is in lowest terms  = 0 = = 3 vertical asymptotes : = , = Horizontal/oblique asymptote: degree o numerator 3 > degree o the denominator, there is no horizontal asymptote degree o numerator 3 = 1 + degree o the denominator, there is an oblique asymptote Oblique asymptote: y = 3 
11 5.3 Sketching the graph o a rational unction 1. Find the domain: i solve q = 0 ii D = { q 0} p q. Find  and yintercepts: yintercept: y = 0  intercepts: numerator = 0 3. Find vertical asymptotes, i any Remark: I = r is ecluded rom the domain and = r is not a vertical asymptote, then the graph o will pass through the point r, reduced r but the point itsel will not be included. We put an open circle around that point The graph o has a hole at = r. Find the horizontal/oblique asymptote, i any. 5. Find the points where the graph crosses the horizontal/oblique asymptote y = m +b i solve the equation = m + b 6. Check or symmetries i I  =, then the graph is symmetric about y ais; ii I  = , then the graph is symmetric about the origin Remark: I the graph is symmetric then only graph unction or >0 and use symmetry to graph the corresponding part or <0 7. Make the sign chart or the reduced i plot intercepts and points ecluded rom the domain on the number line; these points divide the number line into a inite number o test intervals ii choose a point in each test interval and compute the value o at the test point iii based on the sign o at the test point, assign the sign to each test interval Remark: When > 0, then the graph o is above the ais. When < 0, then the graph is below the ais 8. Sketch the graph o using 17: i Draw coordinate system and draw all asymptotes using a dashed line ii plot the intercepts, points where the graph crosses the horizontal/oblique asymptote and the points rom the table in step 7. iii join the points with a continuous curve taking into consideration position o the graph relative to the ais step 7 and behavior near asymptotes.
12 Eample: Graph 1 1 Domain: = 0 = =, =  D = { , } y intercept: y = 0 = 1/= 3  intercepts: + 1 = = 0 =  or = 3 3 Vertical asymptotes = 0 = =  vertical asymptotes: = , = Horizontal/oblique asymptotes Degree o numerator = degree o the denominator, y = 1/1= 1 is the horizontal asymptote 5 Intersection with asymptote: = The graph crosses the horizontal asymptote at = 8, that is at the point 8,1 6 Symmetries: is not the same as , so is not even and thereore not symmetric about yais and are not the same so, is not odd and thereore not symmetric about the origin 7
13 15 positive negative positive.5 negative positive 3 1 8
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