Department of Mathematics, University of Wisconsin-Madison Math 114 Worksheet Sections (4.1),

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1 Department of Mathematics, University of Wisconsin-Madison Math 114 Worksheet Sections (4.1), 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() 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

2 Find the vertical, horizontal and oblique asymptotes of H() = 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

3 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 + = , so we get = 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 f()

4 For (6), 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() So 3 3 > 0 for ( 1, 0) (3, ). 4

5 (b) 3 < Solution: ( + 1) 3( 3) ( 3)( + 1) ( 3)( + 1) + 14 ( 3)( + 1) < Using the zeros of the numerator and denominator, we get: f() + + So 3 < for (, 7) ( 1, 3). 6. Find all the zeros of f() = Solution: a 0 = 6 so factors: 1,, 13, 6 a n = 1 so factor: 1 p = ±1, ±, ±13, ±6. q f(1) = f( 1) = f() = = 0 So = is a zero. Therefore f() = ( )P () where P () is a polynomial of degree. To find P (), we divide by ( ). P () = 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

6 Setting P () = = 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() = 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) = = 8 0 f( 1) = = 0 f() = = 0 So = 1, are zeros. Therefore f() = ( + 1)( )P () = ( )P () where P () is a polynomial of degree. To find P (), we divide 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)( + )( ) Given that = i is a zero of f() = , 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 = ( i)( ( i))( a) = ( i)( + i)( a) = ( + 4)( a) Dividing by + 4, we get a = 4 X 4 X + 4 ) X 3 4X + 4X 16 X 3 4X 6 4X 16 4X

7 So = ( i)( + i)( 4) Therefore the zeros are i, i, and 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)( ) = ( + 1)( 4 + ) =

1. Find the domain of the following functions. Write your answer using interval notation. (9 pts.)

1. Find the domain of the following functions. Write your answer using interval notation. (9 pts.) MATH- Sample Eam Spring 7. Find the domain of the following functions. Write your answer using interval notation. (9 pts.) a. 9 f ( ) b. g ( ) 9 8 8. Write the equation of the circle in standard form given