5.2 Properties of Rational functions

Transcription

1 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

2 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 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, =

3 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 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 : = -, = 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 -

4 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 y-intercepts: y-intercept: 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 1-7: 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.

5 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 y-ais and are not the same so, is not odd and thereore not symmetric about the origin 7

6 1-5 positive negative positive.5 negative positive 3 1 8