Unit 8 - Polynomial and Rational Functions Classwork

Transcription

1 Unit 8 - Polynomial and Rational Functions Classwork This unit begins with a study of polynomial functions. Polynomials are in the form: f ( x) = a n x n + a n 1 x n 1 + a n 2 x n a 2 x 2 + a 1 x + a 0 with a n,a n 1,a n 2,...a 2.a 1,a 0 being real numbers and n is an integer. The degree of the polynomial is n. f ( x) = 4x 2 5x 1 is a polynomial of degree 2. f ( x) = 4 3 x 5 5x 3 x 2 2x 6 is a polynomial of degree 5. 2 f ( x) = 6x 3 is a polynomial of degree 1. f ( x) = 21 is a polynomial of degree 0. 4 We have already studied polynomials of degree 0 (horizontal lines) and polynomials of degree 1 (in the form of y = ax + b), which are linear functions. So we start this section with an analysis of quadratic functions. A. Quadratic Functions If a, b, and c are real numbers with a 0. f ( x) = ax 2 + bx + c is a quadratic function. All quadratic functions graph U shaped curves called parabolas. All parabolas are symmetric with respect to the line called the axis of symmetry. The point where the axis intersects the parabola is called the vertex of the parabola. If the x 2 coefficient a is positive, the graph of the parabola opens upwards as shown in the figure below to the left. If the x 2 coefficient a is negative, the graph of the parabola opens downward as shown in the figure below to the right. The simplest parabolas are in the form y = ax 2. These parabolas have a vertex at the origin. 8. Polynomial and Rational Functions Stu Schwartz

2 The graphs above are y = x 2 and y = x 2 respectively. They are centered at the origin. For y = x 2, as you move one unit right or left, the curve moves one unit up. If you move 2 units to the left or right of the origin, the curve goes 4 units up. 3 units to the left goes 9 units up. The same is true of the y = x 2 but the movement is down and not up. It is good to remember these as the basic parabolas because as we slightly alter the parabola, we will compare the altered one to the basic one. The graph below left compares y = x 2 to y = 2x 2. Note that the graph of y = 2x 2 stretches the basic parabola y = x 2 by a factor of 2. As we go one unit to the left or right, y = 2x 2 goes up 2 units instead of the usual 1. As we go 2 units to the left or right, y = 2x 2 goes up 8 units instead of the usual 4. The graph below right compares y = x 2 to y = 1 2 x 2. Note that the graph of y = 1 2 x 2 stretches the basic parabola y = x 2 by a factor of 1. As we 2 go one unit to the left or right, y = 1 2 x 2 goes up 1 2 unit instead of the usual 1. As we go 2 units to the left or right, y = 1 2 x 2 goes up 2 units instead of the usual 4. We can transform parabolas both left and right and up and down. The transformations will affect the vertex and axis of symmetry. The generalization for these transformations is as follows: Standard Form of Quadratic Equations The quadratic function f x ( ) = a( x h) 2 + k,a 0 is in standard form. The graph of f is a parabola whose axis of symmetry is the line x = h and whose vertex is the point ( h,k). If a > 0, the parabola opens upward and if a < 0, the parabola opens downward. a controls the vertical stretch. If a >1, the parabola is thinner and if a <1, the parabola is fatter. Analyze the following curves and sketch them. Verify by calculator. 1) y = ( x 1) y = ( x + 2) 2 +1 Vertex: Axis: Vertex Axis: 8. Polynomial and Rational Functions Stu Schwartz

3 3. y = ( x 3) 2 4. y = ( x + 4) y = 2( x 1) 2 4 Vertex: Axis: Vertex: Axis: Vertex: Axis: 6. y = 3( x + 3) y = 1 2 x y = 1 ( 2 x 2 ) Vertex: Axis: Vertex: Axis: Vertex: Axis: Unfortunately, quadratic functions are usually written in general form ( y = ax 2 + bx + c). You have to put them into standard form to find out the important information. The technique is called completing the square. To complete the square, move the constant term to the left side, take half of the coefficient of x and square it. Add that amount to both sides. The right side will now be a perfect square and you can factor it. Example: Find the following information for the following: 10. y = x 2 + 6x y = x 2 2x 5 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 8. Polynomial and Rational Functions Stu Schwartz

4 12. y = x 2 x y = x 2 + 9x 6 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 14. y = 2x 2 +12x 15. y = 2x 2 6x 3 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 16. y = 2x 2 + 5x y = 1 2 x 2 + 6x 4 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 8. Polynomial and Rational Functions Stu Schwartz

5 Now we reverse the process. You are given a graph of a quadratic function. You first generate the equation of the function in standard form y = a( x h) 2 + k ( ). [ ] and then change it to general form y = ax 2 + bx + c vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: 8. Polynomial and Rational Functions Stu Schwartz

6 Example 27. Find the equation of the parabola that goes through the points (-3, 11), (4, -3) and (5, 3) and predict the value of y if x were 2. First, on the calculator graph these points using : STAT EDIT and then set up a Plot using 2nd Y =. Finally graph the points (being sure that there is no equations in your Y = ). To graph the points use Zoom 9 : ZoomStat. To find the equation of the parabola, we use this theory: The equation will be in the form y = ax 2 + bx + c and we have 3 points that satisfy this equation. So plug the 3 points into this equation to yield 3 simultaneous equations in 3 unknowns, a, b, and c. (-3, 11) (4, -3) (5, 3) In algebra 2, you learned techniques to solve these equations. We will use the calculator s ability to solve this using matrix theory. First access the matrix menu using 2nd x 1. Choose EDIT and then 1: [A]. You need to change this matrix to a 3 x 3 matrix. Now type in the coefficients to a, b, and c in the equations above: You now need to have another matrix, matrix B which will represent the left side of the equations above. Access the matrix menu again and choose EDIT and then 2: [B] You need to change this matrix to a 3 x 1 matrix. Now type in the three numbers on the left side of the equations. Now for the answer. Mathematically, the solution to the three simultaneous equations in 3 unknowns will be a new matrix [ A] 1 [ B]. So use your matrix menus from the home screen to generate this matrix. 8. Polynomial and Rational Functions Stu Schwartz

7 So your solution is y = x 2 3x 7. Type the equation into Y1 and graph it. It should go all three points. This method will always work for any three points as long as the points are not along a straight line. Finally, we can easily predict the value of the function at x = 2 by plugging 2 into the equation and getting -9. You can also use your VARS Y - VARS 1: Function Y1 command to get the answer as shown on the screen on the right. Find the equation of the parabola going through the following points and determine the y-value for the last point. Show the three simultaneous equations you used. Verify graphically. 28. (0,0), (1, 1), (2, 4) (3, ) 29. (3, -2), (4, 8), (9, 7), (6, ) Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: B. Roots of Polynomial Functions Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: Before we look at roots of polynomial equations, we have to classify all the different types of numbers we encounter in mathematics: Natural numbers (sometimes called counting numbers) 1, 2, 3, 4, Whole numbers (the natural numbers including zero) 0, 1, 2, 3, 4, Integers (positive and negative whole numbers) - -4, -3, -2, -1, 0, 1, 2, 3, 4, Rational numbers numbers that can be written as a fraction a b Repeating decimals are rational. where a and b are both integers and b 0. 3 Irrational numbers numbers that cannot be written as fractions (ex. π, 2, 5, etc). Any number that is a non-repeating decimal is irrational. Real numbers a combination of rational and irrational numbers. All non-imaginary numbers are real. Pure imaginary numbers numbers in the form bi where i = Complex numbers numbers having a real part and an imaginary part in the form of a + bi where a is the real part and bi is the imaginary part. All numbers are complex. The relationship of these numbers is shown by the chart on the next page Polynomial and Rational Functions Stu Schwartz

8 So any number that is an integer is also rational, real, and complex. All numbers are complex and all numbers without an imaginary part are real. All real numbers are either rational or irrational. 1. Classify the following numbers using the following: N Natural, W Whole, I Integer, Rat = Rational, Irr = Irrational, R = Real, PI = Pure imaginary, and C = Complex. a. 12 b. π c d e. -1 f. 3 g. 4i h. 0 5 i. 0.3 j k i l We define a root of a function f ( x) as the value of x which makes the function equal to zero. There are many ways to express the root of a function. The following statements are all equivalent. x = a is a root of f x (a, 0) is an x-intercept of f x ( ) x = a is a zero of f ( x) x = a is a solution of f ( x) = 0 ( ) ( x a) is a factor of f ( x) So, given f ( x) = x 2 4, we set the function equal to zero and get x =2 and x = -2. So we can make the following statements: They all say exactly the same thing. To the right is the graph. x = 2, 2 are roots of f ( x), x = 2, 2 are zeros of f ( x), (2, 0) and (-2, 0) are x- intercepts of f ( x), x = 2, 2 are solutions of f ( x), x 2 is a factor of f ( x), x + 2 is a factor of f ( x). ( ) = x 3 x 2 6x, we set f ( x) = 0 and we get x( x 3) ( x + 2) yielding x = 0, x Given f x = 3, x = -2. So we can make the following statements: They all say exactly the same thing. To the right is the graph. x = 0,3, 2 are roots of f ( x), x = 0,3, 2 are zeros of f ( x), x = 0,3, 2 are solutions of f ( x), (0, 0), (3, 0), and (-2, 0) are x-intercepts of f ( x), x, x 3, and x + 2 are factors of f ( x). 8. Polynomial and Rational Functions Stu Schwartz

9 1 root 2 roots 3 roots infinite # of roots no roots When we graph polynomials, we get smooth graphs that are continuous meaning it has no breaks. a polynomial a polynomial not a polynomial not a polynomial a smooth curve a smooth curve not continuous sharp corners All polynomials have the following characteristics: A polynomial of degree n will have at most n real roots. A polynomial of degree n will have at most n 1 turning points. A turning point is when the curve changes from increasing to decreasing. A quadratic, for example has at most two real roots and at most one turning point. The graph of f x ( ) = x 2 4 as we saw above has exactly 2 roots and one turning point. The graph of f x ( ) = x 2 4 x + 4 has one root and one turning point. A cubic equation can have at most three real zeros. It can have fewer than three but not more than three. It can have at most two turning points. The graph of f x ( ) = x 3 x 2 6xas we saw above has 3 roots and 2 turning points. The graph of f ( x) = x 3 3x 2 + 3x 1 has one zero and no turning points. Polynomials of degree 1 (linear functions): These will be in the form f ( x) = ax + b and we have already studied these. We know that these graph lines. By the generalizations above, we know that lines have at most 1 root and no turning points. Setting ax + b = 0, we get x = b a as the single root. The root can be an integer but will be always be rational (and thus real). 2. Find the roots of the following: a. f ( x) = 5x 10 b. f ( x) = 4x 3 c. y = 8.2x d. y = 4 8. Polynomial and Rational Functions Stu Schwartz

10 Polynomials of degree 2 (quadratic functions): These will be in the form ax 2 + bx + c = 0 and we have already studied these earlier in this unit. These graph parabolas. We focused on the vertex of the parabolas but not the roots. The equation ax 2 + bx + c = 0 is solved by factoring and setting each factor equal to zero. If factorable, the roots are rational (and thus real). 3. Find the roots of the following: Verify using the calculator. a. f ( x) = x 2 2x 15 b. f ( x) = x c. y = 2x 2 + x 12 d. y =10x 2 +17x 20 If a quadratic isn t factorable (and even if it is), you can use the quadratic formula to find the roots. ax 2 + bx + c = 0 x = b ± b2 4ac. b 2 4ac is called the discriminant as it has the ability to discriminate 2a the number and nature of the roots: > real roots b 2 4ac = real root < real roots imaginary roots ( ) If b 2 4ac is a perfect square, your roots are also rational. 4. Find the number and nature of the roots, and the actual roots for the following. Verify using the calculator. a. f ( x) = 40x x 30 b. f ( x) =16x 2 24x + 9 c. f ( x) = x 2 5x 2 d. y = 2x 2 + x + 2 Polynomials of degree 3 (cubic functions): ( ) = x 3 3x 2 6x + 8. By inspection, you see that x = 1 is a root of ( ) ( = 0). But we know that there might be as many as two more solutions. How are they Suppose we wished to find the roots of f x f x found? Since we know that x = 1 is a solution to f ( x), we know that x 1 we divide x 3 3x 2 6x + 8 by x 1, it goes in evenly. ( ) is a factor of x 3 3x 2 6x + 8. That is, if So let s do the division. You did this way back in algebra 1. x 1) x 3 3x 2 6x Polynomial and Rational Functions Stu Schwartz

11 This is quite messy and takes up room. There is a better way. It is called synthetic division. To divide x 3 3x 2 6x + 8 by x 1 synthetically, do the following. We write 1 to the outside because we are dividing by x 1. Draw a vertical line and write the coefficients to x 3 3x 2 6x + 8 to the inside. Then leave a space and draw a line (a): Now drop the first coefficient (b). Multiply the outside number by the bottom number and put it diagonally up from that bottom number (c). Now add that column (d). Repeat the process multiplying the bottom row by the outside number and adding the columns (e) (a) (b) (c) (d) (e) The fact that you ended up with a remainder of zero tells you that x 1 divides evenly into x 3 3x 2 6x + 8. The numbers at the bottom represent the coefficients of the answer. The solution is 1x 2 2x 8. So we are saying that x 3 3x 2 6x + 8 = ( x 1) ( x 2 2x 8). Since we can factor x 2 2x 8 into ( x 4) ( x + 2), we can say that x 3 3x 2 6x + 8 = ( x 1) ( x 4) ( x + 2) and the roots are all rational (and real): x = 1, x = 4 and x = -2. Here is the graph that verifies it. Let s practice synthetic division before we tackle how to solve cubics in general. 5. Do the following division problems synthetically. a. x 3 3x 2 6x + 8 x 2 b. x 3 2x 2 21x 18 x + 3 c. x 3 4x 2 28x 30 x + 2 d. x 3 12x 16 x 4 e. x 4 x 3 22x 2 +16x 96 x 3 f. 2x 3 11x 2 +17x 6 2x 1 The key to solving cubics (or higher degree polynomials) is the fact that is a polynomial is divisible by ( x k), then k is a root. And we know whether or not that polynomial is divisible by ( x k) if it is, when we do synthetic division, we get a remainder of 0. This is summarized by the Factor Theorem: A polynomial f ( x) has a factor ( x k) if and only if f ( k) = 0. In the example above f x from the other direction, f 1 ( ). So by the factor theorem, f 1 ( ) = x 3 3x 2 6x + 8 had a factor of x 1 ( ) = 0. So ( x 1) is a factor of f ( x) = x 3 3x 2 6x + 8. ( ) = 0. Or 8. Polynomial and Rational Functions Stu Schwartz

12 Solving cubics (and higher degree polynomials) is a trial and error process. There are no simple formulas like the quadratic formula to help us. We need to find possible rational roots of a function f ( x) and test them with synthetic division. The Rational Zero Test a.k.a. the p q Test If f ( x) = a n x n + a n 1 x n 1 + a n 2 x n a 2 x 2 + a 1 x + a 0 has integer coefficients, every rational root of f ( x) is in the form: p q where p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n This looks hard but it is actually quite easy. We use it to solve polynomials with powers higher than Find the roots and their nature as well as factoring f ( x) = x 3 5x 2 + 2x + 8 p (all factors of the constant term 8) = q (all factors of the leading coefficient 1) p all possible combinations of q : If f x We do some trial and error. There are 8 possibilities. The best place to start is with the smallest numbers. ( ) = x 3 5x 2 + 2x + 8 is going to factor, then one of these combinations must work (divide evenly). So Try 1: Try -1: -1 works. So that means that x +1 as well the result of the synthetic division. ( ) has to be a factor of f ( x) = x 3 5x 2 + 2x + 8. We know the other factor So we now have that x 3 5x 2 + 2x + 8 = ( x +1) ( x 2 6x + 8). No need for any more trial and error because we can factor x 2 6x + 8. So the final factorization is ( x +1) ( x 2) ( x 4). Our roots are x = -1, x = 2, x = 4. All the roots are rational and real. So the process of solving a cubic is this: 1) Determine all your possible p s. If the leading coefficient is 1, this makes your work easier. q 2) Start trying all your possible p s using synthetic division. All you need is one of them to divide evenly. It q is best to start with the smaller numbers. 3) If k is the number that worked then x k is one of the factors. The other factor is the result of the division. 4) The result of the division is a quadratic. Try and factor it. If it does factor, your roots are all rational. If it doesn t factor, some of your roots will be irrational or imaginary. 8. Polynomial and Rational Functions Stu Schwartz

13 6. For each of the expressions below, find the roots and classify them as well as factoring it. Calculator verify. a. Expression: f ( x) = x 3 8x x 18 Possible Rational Roots: b. Expression: f ( x) = x 3 12x +16 Possible Rational Roots: c. Expression: f ( x) = x 3 + 6x 2 +12x + 8 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

14 d. Expression: f ( x) = x 3 3x 2 3x + 9 Possible Rational Roots: e. Expression: f ( x) = x 3 + 3x 2 + x 2 Possible Rational Roots: f. Expression: f ( x) = x 3 2x 2 + 4x 8 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

15 g. Expression: f ( x) = 2x 3 3x 2 11x + 6 Possible Rational Roots: h. Expression: f ( x) = 4 x 3 + 8x 2 + 5x +1 Possible Rational Roots: i. Expression: f ( x) = 8x 3 4x 2 2x +1 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

16 Polynomials of degree 4 (quartic functions) and higher: The method of solving 4 th degree (and higher) is exactly the same as that of cubics. 1. Develop your possible roots using the p q method. 2. Use synthetic division with your possible roots to find an actual root. If you started with a 4 th degree, that brings down the new problem to solving a cubic. 3. Continue the synthetic division trial process with the resulting cubic. Don t forget that roots can be used more than once. 4. Once you get down to a quadratic, use factoring techniques or the quadratic formula to get the other two roots. a. Expression: x 4 + x 3 7x 2 x + 6 Possible Rational Roots: b. Expression: x 4 4x 3 5x x 36 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

17 c. Expression: x 4 14x 2 12x +16 Possible Rational Roots: d. Expression: x 4 + 2x 3 3x 2 +10x 40 Possible Rational Roots: e. Expression: 4x 4 4 x 3 7x 2 + 8x 2 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

18 f. Expression: x 5 7x 4 +10x 3 + 2x 2 11x + 5 Possible Rational Roots: g. Expression: x 5 5x x 2 80x + 48 Possible Rational Roots: h. Expression: x 5 6x 4 +10x 3 10x 2 + 9x 4 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

19 Working with Imaginary Roots Suppose you were asked to factor/find the roots of f ( x) = x 4 x 3 + 3x 2 4x 4. There are only 6 possible roots: ±1,±2,±4. To save time, let s graph the function. It should be apparent that none of these possible solutions are roots of the function. And without a toehold to whittle the problem down to a quadratic, we are sunk. However, suppose you were told that 2i was a root of the function. Because roots come in pairs (because of the quadratic formula), another root would also be -2i. So let s do synthetic division with first 2i and then -2i. First, a review on imaginary number math: Remember that i 2 = 1. i( 2 i) = 2i i 2 = 2i ( 1) = 2i +1. ( i 4) ( 3i +1) = 3i 2 + i 12i 4 = 3 11i 4 = 7 11i All of this can be easily done on the calculator as shown on the right. So let s continue by doing synthetic division. Remember we are told that 2i and -2i are roots of the function so the remainders must be equal to zero. 2i i i 1 2i 2i So, at this point, we have broken the problem into ( x 2i) ( x + 2i) ( x 2 x 1). x 2 x 1 isn t factorable but has roots 1 ± 5. So we have the two irrational (real) roots 1 ± 5 and the two imaginary roots ±2i. Finally, we 2 2 multiply ( x 2i) ( x + 2i) to get ( x 2 + 4) ( x 2 x 1). Notice that we couldn t have done this problem unless we were told one of the roots. That gives us a toehold to complete the problem. 7. Fully factor and find all of the roots of f ( x) = x 4 4 x 3 +14x 2 36x + 45 given that 2 + i is a root. Steps: Remember if 2 + i is a root, then 2 i is also a root. Do synthetic division with 2 + i you must get a remainder of zero. Do synthetic division with your answer and 2 i you must get a remainder of zero. Either factor your result or use the quadratic formula To present your answer in factored form, use the fact that x 2 + i [ ( )] x ( 2 i) = x 2 Note that f x [ ] = [( x 2) i] [( x 2) + i] ( ) 2 i 2 = x 2 4x + 4 ( 1) = x 2 4x + 5 ( ) has no real roots verified by the graph to the right. 8. Polynomial and Rational Functions Stu Schwartz

20 C. Rational Functions Bottom Heavy and Powers Equal We have spent a good deal of time working with polynomial functions. None of these functions had denominators. We now turn to rational functions in the form of p ( x ) q( x) where p ( x ) and q x functions, q x ( ) 0. For purposes of this document, let s make the following definitions: Bottom-heavy the degree of q( x) is higher than the degree of p( x) Top-heavy - the degree of p( x)is higher than the degree of q( x) Powers-equal the degree of p( x) is the same as the degree of q( x) ( ) are polynomial In this section, we will concentrate on bottom-heavy and powers-equal expressions. Top-heavy expressions will be tackled in section D. Different than section C, where we attempted to find the roots, we are interested in actually graphing the rational functions given. There are a series of steps to follow which will give us the necessary information to graph the function on the supplied graph. Term Definition How to find it a. Domain the set of values that x can take. Usually we express the domain in terms of values x cannot take. Set the denominator equal to zero. The roots of the denominator are what x cannot be. If the denominator is never equal to zero, then the domain is all real numbers. After you do this, factor both numerator and denominator and do any cancellation. From now on, it is b. Vertical Asymptotes (VA) c. Horizontal Asymptote (HA) A vertical line in the form of x = k that the curve approaches but never reaches. VA s are usually shown by a dashed vertical line on your graph. A horizontal line in the form of y = k that the curve approaches the further right or left you go on the graph. Unlike the VA, it is possible for a graph to cross its HA. Usually shown by a dashed horizontal line. (Roots) Where the function crosses the x- axis (if at all). e. Interval Work A sign chart that determines whether the function is positive (above the axis) or negative (below the axis) at critical intervals. best to work with this rational function. Set the denominator equal to zero, VA s are where the denominator equals zero. There can be more than one VA. If the denominator can never equal 0, then the function has no VA. Rational functions always has an HA. If the function is bottom heavy, the HA is y = 0. If the function is powers-equal the HA will be coefficient of numerator's highest power y = coefficient of denminator's highest power Set the numerator equal to zero. If it cannot equal zero, then there are no zeros and the function never crosses the x-axis. Your critical values are any x-values you found in b (VA) and d (Zeros) above. Make a number line with those critical values marked. Then determine the sign of the function in the intervals created by plugging in sample numbers to the function. Set x equal to zero. f. y-intercept Where the function crosses the y- axis, if at all. To graph the function, use the sign chart to determine whether the graph approaches the VA from the high or low side. Then it s a matter of playing connect the dots. 8. Polynomial and Rational Functions Stu Schwartz

21 Examples: Graph the following. 1. y = x 4 x + 2 c. HA : 2. y = 4 x 3 c. HA : 3. y = x 2 + 3x x 2 + 5x + 6 c. HA : x 4. y = x 2 16 c. HA : 8. Polynomial and Rational Functions Stu Schwartz

22 5. y = x 2 16 x 2 4 c. HA : 6. y = x 2 9 x c. HA : 7. y = x x 2 c. HA : 8. y = 2x 2 2 x 2 3x c. HA : 8. Polynomial and Rational Functions Stu Schwartz

23 D. Rational Functions Top Heavy When rational functions are top heavy (higher power in the numerator), there is one major difference in the shape of the graphs. Instead of a horizontal asymptote, the function has an oblique asymptote (OA) or sometimes called a slant asymptote. An OA operates like a horizontal asymptote except that instead of being horizontal, the line is on a slant. Below are some graphs with OA s which are shown with dashed lines. It is also possible for an OA to be not a line but a curve as shown in the last graph. one VA and one OA two VA s and one OA one VA and one OA (a parabola) To find the OA, you divide the denominator into the numerator. It will not go in evenly. The OA is the quotient without the remainder. In the case of linear denominators, you can do the long division by synthetic division. For the following expressions, find the requested information and graph using the calculator to verify. 1. y = x 2 x 6 x 1 Synthetic/Long Division : c. OA : 2. y = x 2 + 3x x + 2 Synthetic/Long Division : c. OA : 8. Polynomial and Rational Functions Stu Schwartz

24 3. y = x 3 x 1 Synthetic/Long Division : c. OA : 4. y = x 3 + x 2 2 x +1 Synthetic/Long Division : c. OA : 5. y = x 3 4x 2 11x + 30 x 2 16 Synthetic/Long Division : c. OA : 8. Polynomial and Rational Functions Stu Schwartz

25 E. Partial Fraction Decomposition 5 You are used to adding two fractions like x You find an LCD and multiply each term by the x x + 3 missing factor. + 2 x 1 5x x 2 7x +13 = = x 1 x + 3 x + 3 x 1 x 1 x 2 + 2x 3. ( )( x + 3) 7x +13 However, at times, there is a necessity to reverse the process: to change x 2 + 2x 3 to 5 x This is x + 3 called partial fraction decomposition. The method is called the Heaviside Method. (Oliver Heaviside -1850). The way to begin is to factor the denominator: 7x +13 x 2 + 2x 3 = 7x +13 x 1 ( )( x + 3) Realize that if this were to be written as the sum of two fractions, there must be two numbers A and B such that 7x +13 x 1 x 1 + B x + 3. ( )( x + 3) = A To find A, take your finger and cover up the ( x 1) in Now let x = 1 and plug that into 7x +13 x + 3 ( ) To find B, take your finger and cover up the ( x + 3) in Now let x = -3 and plug that into Hence 7x +13 ( x 1) ( x + 3) = A x 1 + 7x +13 x 1 ( ) B x + 3 = 5 x x + 3 Write the partial fraction decomposition for: 7x +13 ( x 1) ( x + 3). It will look like this: 7x +13 ( x + 3) and you get A = 20 4 = 5 7x +13 ( x 1) ( x + 3). It will look like this: 7x +13 ( x 1) and you get B = 8 4 = 2 1) x 19 x 2 + 4x 5 2) 5x 2 4x 4 x 3 4x 8. Polynomial and Rational Functions Stu Schwartz

26 This method of partial fraction decomposition will not work if the denominator has any repeated factors. For 2x 2x instance, to perform partial fraction decomposition on, we factor the denominator and get x 2 4x + 4 x 2 Since we don t have distinct linear terms in the denominator, the Heaviside method doesn t work. Instead we write: 2x ( x 2) = A 2 x 2 + B ( x 2) Now as before, let x = 2 and we get 4 = B To find A, let x equal any number other than 2. So, if x = 3, then we get 6 = A + 4 and A = 2. We now multiply out by the LCD ( 2 x 2 )2 to get 2x = A( x 2) + B ( ) 2. So 2x ( x 2) = 2 2 x x 2 ( ) 2 Perform partial fraction decomposition on 3) 12x + 43 x 2 + 8x +16 4) 5x 2 12x + 3 x 3 2x 2 + x 8. Polynomial and Rational Functions Stu Schwartz

27 Unit 8 - Polynomial and Rational Functions Homework 1. Analyze the following curves and sketch them. Verify by calculator. a. y = ( x +1) 2 b. y = ( x +1) 2 c. y = x 2 5 Vertex: Axis: Vertex: Axis: Vertex: Axis: d. y = 3 x 2 e. y = ( x 2) 2 +1 f. y = ( x + 4) Vertex: Axis: Vertex: Axis: Vertex: Axis: g. y = 2x 2 5 h. y = 1 ( 2 x 2 ) 2 i. y = 2( x +1) Vertex: Axis: Vertex: Axis: Vertex: Axis: 8. Polynomial and Rational Functions Stu Schwartz

28 j. y = 3( x +1) 2 5 k. y = 4 1 ( 2 x 2 ) 2 l. y = 2.5 ( x + 2.5) 2 Vertex: Axis: Vertex: Axis: Vertex: Axis: 2. Find the following information for the following: a. y = x 2 4x b. y = x 2 8x 18 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider c. y = x 2 3x d. y = x 2 + 5x 1 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 8. Polynomial and Rational Functions Stu Schwartz

29 e. y = 2x 2 20x +15 f. y = 2x 2 18x + 7 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider g. y = x 2 +10x + 4 h. y = 1 2 x 2 + 8x + 22 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider i. y = 1 2 x 2 + 5x j. y = 2 3 x 2 +12x + 61 Horizontal shift: Horizontal Shift: Vertical shift: Vertical Shift: Vertex: Vertex: Axis: Axis: Open: Up Down Open: Up Down Compare to y = x 2 Narrower Wider Compare to y = x 2 Narrower Wider 8. Polynomial and Rational Functions Stu Schwartz

30 3. You are given a graph of f ( x). Find the equation of the parabola in standard form and then general form. a. b. c. vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: d. e. f. vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: g. h. i. vertex: a = vertex: a = vertex: a = Standard form: Standard form: Standard form: General form: General form: General form: 8. Polynomial and Rational Functions Stu Schwartz

31 4. Find the equation of the parabola going through the following points and determine the y-value for the last point. Show the three simultaneous equations you used. Verify graphically. a. (3,1), (6, 43), (8, 91) (4, ) b. (2, -14), (4, 10), (5, 10.5), (0, ) Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: c. (8,-120), (10, 212), (15, 0) (9, ) d. (2, -14), (26, 46), (34, 66), (-5, ) Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: Equation 1: Equation 2: Equation 3: Parabola: y = Predicted y-value: 5. Classify the following numbers using the following: N Natural, W Whole, I Integer, Rat = Rational, Irr = Irrational, R = Real, PI = Pure imaginary, and C = Complex. a. 2 i b c. 1 2 d e f. 64 g h. 2 2 i. i 2 j. 1 π k. 0 l. π π 0 6. Find the number and nature of the roots, and the actual roots for the following. Verify using the calculator. a. f ( x) =12x 25 b. f ( x) = 29 c. f ( x) = x 2 9x 10 d. f ( x) = 4x 2 1 e. f ( x) = x 2 + 5x 8 f. f ( x) = 9x x + 25 g. f ( x) = x h. f ( x) =18x 2 + x 4 i. f ( x) = 3x 2 2x 3 j. f ( x) = 2x 2 + 5x 3 k. f ( x) = 5x 2 x + 2 l. f ( x) = x Do the following division problems synthetically. 8. Polynomial and Rational Functions Stu Schwartz

32 a. x 3 6x 2 +11x 6 x 2 b. x 3 7x 2 6x + 72 x + 3 c. x 3 15x x 120 x 5 d. x x + 4 e. x 4 8x 3 3x x 200 x 5 f. 2x 3 5x 2 11x 4 2x For each of the expressions below, find the roots and classify them as well as factoring it. Calculator verify. a. Expression: f ( x) = x 3 6x 2 + 3x +10 Possible Rational Roots: b. Expression: f ( x) = x 3 x 2 8x +12 Possible Rational Roots: c. Expression: f ( x) = x 3 + 2x 2 8x 16 Possible Rational Roots: d. Expression: f x ( ) = x 3 3x 2 + 3x 9 Possible Rational Roots: 8. Polynomial and Rational Functions Stu Schwartz

33 e. Expression: f ( x) = x 3 + x 2 + 2x 4 Possible Rational Roots: f. Expression: f ( x) = 3x 3 5x 2 47x 15 Possible Rational Roots: g. Expression: f ( x) =12x 3 + 8x 2 x 1 Possible Rational Roots: Want more practice? Do these on notebook paper. h. f ( x) = x 3 11x x 40 i. f ( x) = x 3 13x x 80 j. f ( x) = x 3 +12x x Polynomial and Rational Functions Stu Schwartz

34 k. f ( x) = x 3 6x 2 5x + 30 l. f ( x) = x 3 5x 2 + 7x 2 m. f ( x) = x x 2 + 9x +10 n. f ( x) = 8x 3 +18x 2 17x + 3 o. f ( x) =16x x 2 + 8x +1 p. f ( x) = 4 x 3 + 3x 2 +11x 3 9. For each of the expressions below, find the roots and classify them as well as factoring it. Calculator verify. a. Expression: x 4 11x 3 13x 2 +11x +12 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary b. Expression: x 4 x 3 9x 2 11x 4 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary c. Expression: x 4 2x 3 20x x + 20 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary d. Expression: x 4 3x 3 +12x 2 30x + 20 Possible Rational Roots: Roots 8. Polynomial and Rational Functions Stu Schwartz

35 Classify each Rat l Irrat l Real Imaginary e. Expression: 16x 4 32x x 2 8x +1 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary f. Expression: x 5 12x x 3 90x x 24 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary g. Expression: x 5 5x 4 + 8x 3 8x 2 +16x 16 Possible Rational Roots: Roots Classify each Rat l Irrat l Real Imaginary 8. Polynomial and Rational Functions Stu Schwartz

36 10. For the following functions, fully factor and find all roots with the given information. a. f ( x) = x 4 x 3 x 1, i is a root b. f ( x) = x 4 x 3 +11x 2 9x +18, 3i is a root c. f ( x) = x 4 + 2x 3 5x 2 14x 14, 1+ i is a root d. f ( x) = x 4 4x 3 +10x 2 8x +16, 2 2i is a root 11. For each of the following rational functions, find the requested information and sketch. Verify using the calculator. a. y = 3 x + 2 c. HA : b. y = 3x x + 2 c. HA : 8. Polynomial and Rational Functions Stu Schwartz

37 x c. y = x 2 + 2x 3 c. HA : d. y = x 2 x 2 16 c. HA : e. y = x 2 3x 4 x 2 c. HA : f. y = x 2 2x +1 x 2 + 2x 8 c. HA : 8. Polynomial and Rational Functions Stu Schwartz

38 4x g. y = x 2 x 12 c. HA : h. y = 4x + 8 x c. HA : i. y = x x 2 +1 c. HA : j. y = 1 x x x + 2 (hint: get a common denominator) c. HA : 8. Polynomial and Rational Functions Stu Schwartz

39 12. For each of the following rational functions, find the requested information and sketch. Verify using the calculator. a. y = x 2 5x 6 x 3 Synthetic/Long Division : c. OA : b. y = x 2 + 2x +1 x + 2 Synthetic/Long Division : c. OA : c. y = x x 2 x +1 Synthetic/Long Division : c. OA : 8. Polynomial and Rational Functions Stu Schwartz

40 d. y = x 3 x 1 Synthetic/Long Division : c. OA : e. y = x 3 2x 2 4x + 8 x 1 Synthetic/Long Division : c. OA : f. y = x 3 3x + 2 x 2 2x Synthetic/Long Division : c. OA : 8. Polynomial and Rational Functions Stu Schwartz

41 13. Perform partial fraction decomposition on the following. a) 5x + 55 x 2 + 5x b) 3x 57 x 2 8x + 7 c) 2x x 12 x 3 + x 2 12x d) x 2 +15x 46 x 3 + x 2 10x + 8 e) 3x +16 x 2 +12x + 36 f) 4x 2 7x 100 x 3 10x x 8. Polynomial and Rational Functions Stu Schwartz