Department of Mathematics, University of Wisconsin-Madison Math 114 Worksheet Sections (4.1), 4.-4.6 1. Find the polynomial function with zeros: -1 (multiplicity ) and 1 (multiplicity ) whose graph passes through the point (, 4). Solution: We have f() = a( + 1) ( 1) with f( ) = 4 f( ) = a( + 1) ( 1) 4 = 9a a = Answer: f() = ( 1) ( + 1). For function f() = ( 3)( + 4), (1) determine the end behavior of the graph of the function, () find the - and y- intercepts of the graph of the function, (3) determine the zeros and their multiplicity, use the information to determine whether the graph crosses or touches the -ais at each -intercept, (4) determine the maimum number of turning points on the graph of the function, () Use the information from (1) to (4) to draw a complete graph of the function. Solution: For (1), Since the highest degree term is 4, we have f() 4. For (), -intercept: (0, 0), (3, 0), ( 4, 0). y-intercept: (0, 0). For (3), = 0, multiplicity. = 3 and = 4 multiplicity 1. So f() crosses the -ais at = 3 and at = 4, f() touches -ais at = 0. f() + + 4 0 3 For (4), since f() has degree 4, it has at most 3 turning points and since f() touches the -ais at = 0 and must cross the -ais at = 4 and at = 3, it has at least 3 turning points, so it must have 3 turning points. For (), 1
40 0 4 4 0 40 60 3. Find the vertical, horizontal and oblique asymptotes of H() = 3 8 + 6. Solution: Vertical asymptotes: 3 8 H() = + 6 = ( )( + + 4). We need to investigate what is happening when ( )( 3) the denominator equals 0: When goes close to, the value of the function goes close to a finite number, it does not go to infinity, so there is no asymptote at = (there is a hole). When goes close to 3, the value of the function become very high (on the right of 3) or very low (on the left of 3) so there is a vertical asymptote at = 3. Horizontal asymptotes: There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Oblique asymptotes: if becomes large, H() 3 =, so y = + b. To find b, we do long division. X + X X + 6 ) X 3 8 X 3 + X 6X X 6X 8 X + X 30 19X 38
19 38 The whole process could also be done entirely by long division. The remainder + 6 goes close to zero when becomes very large so the function looks more and more like y = + therefore the oblique asymptote is y = +. ( 1) 4. Follow the steps 1 through 6 to analyze the graph of R() = ( + 3) 3. (1) Find the domain of the function. () Find the intercepts of the graph and determine its multiplicity. (3) Find vertical asymptotes and determine the behavior of R on either side of the asymptotes. (4) Find horizontal or oblique asymptote. () Use the zeros of the numerator and denominator of R to divide the -ais into intervals. Determine where the graph of R is above or below the -ais by choosing a number in each interval and evaluation R there. (6) Use the results obtained in steps through 6 to graph R(). Solution: For (1), the domain is 3. For (), -intercept: (0, 0), (1, 0). y-intercept: (0, 0). Since = 0 has multiplicity 1, R crosses the -ais at = 0, and since = 1 has multiplicity, the graph touches the -ais at = 1. For (3), vertical asymptotes is: = 3, and the function goes to positive infinity when approaches to 3 from the left side and to negative infinity when approaches to 3 from the right side. For (4), horizontal asymptote: y = 1 (the leading term in both numerator and denominator is 3 ). To find the intersections of the graph with the asymptote, set R() = 1, we get ( 1) = (+3) 3, multiplying out, we get 3 + = 3 + 9 + 7 + 7, so we get 11 + 6 + 7 = 0, and there is no real solution, so the graph doesn t intersect the asymptote. For (),since R( 4) = 100,on (, 3) R is positive, similarly, R( 1) = 1, so on ( 3, 0) R is negative, and R( 1 ) = 1, so on (0, 1) R is positive, R() =, so on (1, + ) R is positive. 343 1 f() + + + 3 0 1 3
For (6), 00 4 3 1 1 00. Solve each inequality algebraically. (a) 3 3 > 0. Solution: 3 3 > 0 ( 3) > 0 ( 3)( + 1) > 0 Using the zeros of y = 3 3 = ( 3)( + 1), we get: f() + + 1 0 3 So 3 3 > 0 for ( 1, 0) (3, ). 4
(b) 3 < 3 + 1. Solution: 3 3 3 + 1 ( + 1) 3( 3) ( 3)( + 1) + 3 + 9 ( 3)( + 1) + 14 ( 3)( + 1) < 3 + 1 Using the zeros of the numerator and denominator, we get: f() + + So 3 < 3 + 1 7 1 3 for (, 7) ( 1, 3). 6. Find all the zeros of f() = 3 8 + 6. Solution: a 0 = 6 so factors: 1,, 13, 6 a n = 1 so factor: 1 p = ±1, ±, ±13, ±6. q f(1) = 1 8 + 6 0 f( 1) = 1 8 6 6 0 f() = 8 3 + 0 6 = 0 So = is a zero. Therefore f() = ( )P () where P () is a polynomial of degree. To find P (), we divide 3 8 + 6 by ( ). P () = 6 + 13. The result of the long division is X 6X + 13 X ) X 3 8X + X 6 X 3 + X 6X + X 6X 1X 13X 6 13X + 6 0
Setting P () = 6 + 13 = 0, we get = 6 ± 36 4(1)(13) = 6 ± 16 So the zeros are, 3 + i, and 3 i. = 3 ± i 7. Find the rational zeros of f() = 4 3 6 + 4 + 8 and use the zeros to factor the polynomial over the real numbers. Solution: a 0 = 8 so factors: 1,, 4, 8 a n = 1 so factor: 1 p = ±1, ±, ±4, ±8. q f(1) = 1 1 6 + 4 = 8 0 f( 1) = 1 + 1 6 4 + 8 = 0 f() = 16 8 4 + 8 + 8 = 0 So = 1, are zeros. Therefore f() = ( + 1)( )P () = ( )P () where P () is a polynomial of degree. To find P (), we divide 4 3 6 + 4 + 8 by ( ). The result of the long division is P () = 4. X 4 X X ) X 4 X 3 6X + 4X + 8 X 4 + X 3 + X 4X + 4X + 8 4X 4X 8 So f() = ( + 1)( )( 4) = ( + 1)( )( )( + ) = ( + 1)( + )( ). 0 8. Given that = i is a zero of f() = 3 4 + 4 16, find the remaining zeros. Solution: If i is a zero, i is also a zero. Let a be the third zero. So we could write f() as 3 4 + 4 16 = ( i)( ( i))( a) = ( i)( + i)( a) = ( + 4)( a) Dividing 3 4 + 4 16 by + 4, we get a = 4 X 4 X + 4 ) X 3 4X + 4X 16 X 3 4X 6 4X 16 4X + 16 0
So 3 4 + 4 16 = ( i)( + i)( 4) Therefore the zeros are i, i, and 4. 9. Find a polynomial function of degree 4 with zeros: 1 (multiplicity ) and +i. Solution: Let s pick a 4 = 1. If + i is a zero, i is also a zero. We have f() = ( 1)( 1)( ( + i))( ( i)) = ( + 1)( i)( + i) = ( + 1)[( ) (i) ] = ( + 1)( 4 + 4 + 1) = ( + 1)( 4 + ) = 4 6 3 + 14 14 + 7