Transcript Chapter 3- Polynomial and Rational Functions
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How does a mountain range relate to polynomials and their graphs?
Definition: If f is a polynomial function and f(a) ≠f(b) for a ◦ Step 1 Identify a and b values ◦ Step 2 Plug a and b into f(x) and solve ◦ Step 3 If f(a) and f(b) have opposite signs, you know that f(c)=0 for at least one real number between a & b ◦ ◦ ◦ Using the Intermediate Value Theorem ◦ Show that f(x)= x⁵+2x⁴-6x ᶟ +2x-3 has a zero between 1 and 2. Step 1: identify a and b a=1, b=2 Step 2: solve f(a) and f(b) f(a)= f(1)= 1+2-6+2-3=-4 f(b)=f(2)=32+32-48+4-3=17 Step 3: do f(a) and f(b) have opposite signs? f(a)=-4 and f(b)=17, yes! Conclusion: there is a c where f(c)=0 between 1 and 2 ◦ ◦ ◦ Sketching graphs of degree greater than 2 Step 1 Find the zeros Step 2 Create a table showing intervals of positive or negative signs for f(x). Step 3 Find where f(x)>0 and where f(x)<0 Construct a table showing intervals of positive or negative values of f(x) Interval Sign of (x-1) , 1 ( 1 , 4 ) + ( 4 , ) + Sign of (x-4) Sign of f(x) Position of graph + Above x axis Below x axis + + Above x axis • , 1 and ( 4 , ) • ( 1 , 4 ) ◦ ◦ Sketching the graph of a polynomial function of degree 3 Ex f ( x ) x 3 x 2 4 x 4 Step 1: Find Zeros Group terms f ( x ) ( x 3 x 2 ) ( 4 x 4 ) ◦ Factor out x² and -4 ◦ Factor out (x+1) f ( x ) x 2 ( x 1 ) 4 ( x 1 ) f ( x ) ( x 2 4 )( x 1 ) ◦ Difference of squares f ( x ) ( x 2 )( x 2 )( x 1 ) ◦ Therefore, the zeros are -2, 2, and -1 Step 2: Create a Table Interval Sign of x+2 Sign of x+1 Sign of x-2 Sign of f(x) Position of graph ( , 2 ) ( 2 , 1 ) ( 1 , 2 ) ( 2 , ) f(x)>0 if x is in ( f(x)<0 if x is in ( 2 , , 1 ) 2 ) ( ( 2 , 1 , ) 2 ) My version of graph (there are multiple ways to do this graph) y x Ex Find the intervals ◦ ◦ f(x) is above x-axis when x is in y ( 0 , 2 ) ( 3 , 1 ), ( 1 , 0 ), and ( 2 , ) x ◦ ◦ ◦ ◦ Steps Step 1: Assign f(x) to Y1 on a graphing calculator Step 2: Set x and y bounds large enough to see from [-15,-15] and [-15,15] This allows us to gauge where the zeros lie from a broad perspective Step 3: Readjust bounds once you know where zeros are more likely to be found Step 4: Use zero or root feature on calculator to estimate the real zero Ex/ Estimate the real zeros of f ( x ) x 3 4 . 6 x 2 5 . 72 x 0 . 656 Graph this function as Y1 in calculator Set bounds as [-15,15] by [-15,15]. Readjust to where zeros might exist, you may use [-1,3] by [-1,3] Find actual root by using zero or foot feature on your calculator Actual root is 0.127 If g(x) is a factor of f(x), then f(x) is divisible by g(x). For example, is divisible by each of the following . However, we can use Polynomial Long Division to find a quotient and a remainder. We will first divide 178 by 8 to find the quotient and remainder. Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing). Step 2: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol. Step 3: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol. Step 4: Subtract and bring down the next term. Step 5: Repeat Steps 2, 3, and 4 until there are no more terms to bring down. Step 6: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer. http://www.youtube.com/watch?v=smsKMWf8ZCs x 2 3 x 1 x 4 0 x 3 0 x 2 0 x 81 x 2 3 x 1 x 4 0 x 3 0 x 2 x 2 0 x 81 x 4 3 x 3 x 2 x 2 3 x 1 x 4 0 x 3 x 2 0 x 2 0 x 81 x 4 3 x 3 x 2 3 x 3 x 2 0 x x 2 3 x 1 x 4 0 x 3 x 2 0 x 2 3 0 x x 81 x 4 3 x 3 x 2 3 x 3 x 2 0 x 3 x 3 9 x 2 3 x 8 x 2 3 x 81 Once we have a quotient and remainder, we must write our final answer.. x 2 3 x 1 x 4 0 x 3 0 x 2 x 2 0 3 x x 8 81 x 4 3 x 3 x 2 3 x 3 x 2 0 x 3 x 3 9 x 2 3 x 8 x 2 3 x 81 8 x 2 24 x 8 21 x 89 Final Answer: x 2 x 4 81 3 x 1 x 2 3 x 8 21 x 89 x 2 3 x 1 2 x 3 2 x 3 7 x 2 2 x 9 2 x 3 2 x 3 x 2 7 x 2 2 x 2 x 2 9 2 x 3 3 x 2 4 x 2 2 x 4 x 2 6 x 4 x 9 4 x 6 15 Solution : 2 x 3 7 x 2 x 2 3 2 x 9 x 2 2 x 2 15 2 x 3 x 2 x 6 x 3 3 x 2 4 x 12 x 2 x 6 x 3 3 x 2 4 x x 2 12 x 3 x 2 6 x 2 x 2 2 x 12 2 x 2 2 x 12 0 Solution : x 3 3 x 2 x 2 4 x 12 x 6 x 2 x If a polynomial f(x) is divided by x-c, the the remainder is f(c). If Example: f ( x ) x 3 2 x 3 3 x 2 x 3 x 2 5 x 5, use the remainder x 2 x 3 theorem to find x 2 3 x 2 x x 1 5 f(-2). x 3 2 x 2 Therefore, f (-2) 3. x 2 x x 2 2 x x 5 x 2 3 A polynomial f(x) has a factor x-c if and only if f(c)=0. Example: Show that x 2 is a factor of f ( x ) x 3 4 x 2 3 x 2. Find a polynomial f(x) of degree 3 that has zeros 3, -1, and 1. pg 236 (# 1,2,8,17-20) http://www.mesacc.edu/~scotz47781/mat12 0/notes/divide_poly/long_division/long_divis ion.html Synthetic Division is the “shortcut” method for dividing polynomials. Let' s compare dividing x 4 x 3 2 x 2 by x 1 using both long division and synthetic Long Division x 3 2 x 2 2 x 4 division. x 1 x 4 x 3 0 x 2 2 x 2 x 4 x 3 2 2 x x 3 3 0 x 2 2 x 2 2 2 x x 2 2 2 2 x x 4 x 4 x 2 4 1 Synthetic Division 1 1 1 1 2 0 2 2 2 2 4 x 3 2 x 2 2 x 4 2 4 6 6 Use synthetic division to find the quotient q(x) and remainder r if f ( x ) 4 x 4 x 3 3 x 2 5, is divided by x 3 . ◦ Using synthetic division to find zeros. What must we show for a value to be a zero of f(x)? Think Factor Theorem… Show that 11 is a zero of f ( x ) x 3 8 x 2 29 x 44. Use synthetic division to find f(3) if f ( x ) 3 x 4 x 3 3 x 2 x 15 ◦ ◦ ◦ You should now recognize the following equivalent statements. If f(a)=b, then: 1. The point (a,b) is on the graph of f. 2. The value of f at x=a equals b; ie f(a)=b. 3. If f(x) is divided by x-a, then the remainder is b. ◦ ◦ ◦ ◦ Additionally, if b=0 then the following are also equivalent. 1. The number a is a zero of the function f. 2. The point (a,0) is on the graph of f; a is an x-int. 3. The number a is a solution of the equation f(x)=0. 4. The binomial x-a is a factor of the polynomial f(x). pg 237 (# 21-29 odd,32,35,38) If a polynomial f(x) has positive degree and complex coefficients, then f(x) has at least one complex zero. If f(x) is a polynomial of degree n>0, then there exist n complex numbers such that f ( x ) a ( x c 1 )( x c 2 )...( x c n ), Where a is the leading coefficient of f(x). Each number is a zero of f(x). A polynomial of degree n>0 has at most n different complex zeros. f f f Find the zeros of the polynomial, state the multiplicity of each zero, find the y-int, and sketch the graph. f ( x ) 1 25 ( x 2 )( x 3 ) 3 ( x 1 ) 2 Find a polynomial f(x) in factored for that has degree 3; has zeros 3, 1, and -1; and satisfies f(-2)=3. If f(x) is a polynomial of degree n>0 and if a zero of multiplicity m is counted m times, then f(x) has precisely n zeros. Exactly how many zeros exist for the following polynomial? f ( x ) 3 ( x 2 )( x 3 ) 3 ( x 1 ) 2 ( x 2 ) 5 pg 249 (# 1-7 odd, 11-15 odd, 19-23 odd) If f(x) is a polynomial with real coefficients and a nonzero constant term, then: ◦ 1. The number of positive real zeros of f(x) either is equal to the number of variations of sign in f(x) or is less than that number by an even integer. ◦ 2. The number of negative real zeros of f(x) either is equal to the number of variations of sign in f(-x) or is less than that number by an even integer. We can use the N-Zeros Thm to determine the total number of zeros possible. Use Descartes’ Rule of Signs to determine the total possible number of positive and negative zeros (and all lesser combinations). Any unaccounted for zeros must then be imaginary zeros. Find the total number of zeros possible for f(x)=0 where f ( x ) 3 x 5 2 x 4 3 x 3 6 x 5 Number of positive solutions Number of negative solutions Number of imaginary solutions Total number of solutions ◦ Suppose that f(x) is a polynomial with real coefficients and a positive leading coefficient and that f(x) is divided synthetically by x-c. Then: 1. If c>0 and if all numbers in the third row of the division process are either positive or zero, the c is a upper bound for the real zeros of f(x). ◦ 2. If c<0 and if the numbers in the third row of the division process are alternately positive and negative (a 0 can be counted as positive OR negative), then c is a lower bound for the real zeros of f(x). Determine the smallest and largest integers that are upper and lower bounds for the real solutions of f(x)=0 if f ( x ) x 4 x 3 2 x 2 3 x 6 Suppose is a polynomial with real coefficients. All of the real zeros of f(x) are in the interval ( M , M ) , wher e M max a n , a n 1 ,..., a 1 , a 0 1 a n Determine the interval in which all of the real solutions of f(x)=0 exist if f ( x ) 3 x 5 2 x 4 3 x 3 6 x 5 Finding Zeros Practice WS (# 1-9) Quiz Tomorrow If a polynomial f(x) of degree n>1 has real coefficients and if with b≠0 is a complex zero of f(x), then the conjugate . is also a zero of f(x). Find a polynomial of degree 3 with real coefficients where 2 and (3-i) are zeros of the polynomial. Every polynomial with real coefficients and positive degree n can be expressed as a product of linear and quadratic factors with real coefficients such that the quadratic factors are irreducible over ℝ. If the polynomial f ( x ) a n x n a n 1 x n 1 ... a 1 x a 0 has integer coefficients and if c/d is a rational zero of f(x) such that c and d have no common prime factor, then: 1. the numerator, c, of the zero is a factor of the constant term 2. the denominator d of the zero is a factor of the leading coefficient All possible rational zeros of a polynomial are of the form: Possible Rational Zeros factors of the constant t erm factors of the leading coefficien a 0 t a n Find all possible rational zeros using rational zeros theorem. Then factor and find any remaining zeros. f ( x ) 3 x 4 2 x 3 x 2 12 x 4 pg 259 (# 7,18,19,23,24,29) Top Shelf pg 14 (#1,2,5,6,8; Due 11/11 ) Prentice Hall p 46 Fundamental Thm Algebra WS (# 1-4,8,9,10,13,17) A function f is a rational function if f ( x ) g ( x ) h ( x ) where g(x) and h(x) are polynomials. The domain of f consists of all real numbers except the zeros of the denominator h(x) . Find the domain of the following rational functions: f ( x ) 1 2 x 3 g ( x ) x 2 4 x 1 3 x 2 g ( x ) x 3 x 2 27 9 D : , 3 2 3 2 , D : , 2 2 , 1 1 , D : Simplify and graph the following rational expression: f ( x ) x 2 x 5 x 2 6 Definition: The line x=a is a vertical asymptote for the graph of the function f if f ( x ) or f ( x ) as x a or x a as x approaches right. a from either the left of the Create an xy-table and look at the values of f(x) as x approaches the vertical asymptote. f ( x ) 1 x 1 Definition: The line y=c is a horizontal asymptote for the graph of the function f if f ( x ) c as x or x Let f be a rational function given by where f x n n d and x d 1. When n m , the a n b m x x n m a n b m 1 1 x x n 1 m 1 a 1 b 1 x x a 0 b 0 are polynomial x axis y 0 is s with the no common horizontal factors. asymptote. 2. When n m , the line y a b m n (ratio of leading coeffients ) is the horizontal asymptote. 3. When n m , there is no horizontal asymptote. Let f be a rational function given by f where n x a n b m x x n a n b m 1 1 x x n n d and x d m m are polynomial 1 1 s with a 1 b 1 no x x a 0 b 0 common factors. 1. When the degree of the degree of the denominato numerator r, y 0 is is less the than the horizontal asymptote. 2. When the degree of the denominato r is equal to the degree of the numerator, the line y a b m n (ratio of leading coeffients ) is the horizontal asymptote. 3. When the degree of the numerator is greater than the degree of the denominato r, there is no horizontal asymptote. If a factor can be reduced that would cause a vertical asymptote there will be a hole instead of a vertical asymptote! f ( x ) x 2 x 2 1 5 x 4 f ( x ) x 14 5 x 10 f ( x ) 3 x 2 x 2 1 9 f ( x ) x 2 x 1 5 x 6 f ( x ) x 2 x 1 5 x 6 ◦ ◦ pg 276 (# 1,2,3,5,6,8,11,15) Only find the Domain, VA, HA, and reduced form of the rational function. We will graph them tomorrow! Assume that f where g and h g h a n x n b m x m a n 1 b m 1 x x n m 1 1 a 1 b 1 x x a 0 b 0 are polynomial s with no common factors. 1. Find the x intercepts , i.e. all real zeros of g(x) . Plot the correspond ing points on the x axis. 2. Find the real zeros of the denominato r h(x) . For each real zero a , sketch the vertical asymptote x a with a dashed line. 3. Find the y intercept f( 0 ) , if it exists, and plot the point ( 0 ,f( 0 )) . 4. Apply the theorem on horizontal asymptotes . If there is a horizontal asymptote y c , sketch it with a dashed line. 5. If there is a HA y c , determine whether it intersects the graph. The x coordinate s of the points of intersecti on are the solutions to the equation f(x) c . Plot these points, if they exist. 6. Sketch the graph of f in each of the regions in the xy plane determined by the VAs in guideline 2. If necessary, use the sign of specific function value s to tell whether t he graph is above or below the x axis or the HA. Use guideline 5 to decide whether t he graph approaches the HA from above or below. f ( x ) 4 x 1 2 x 3 f ( x ) x 2 x 1 2 x 3 f ( x ) x 2 x 2 3 x 25 4 pg 46 Dennison PreCalc (find VA, HA, SA, y int, graph # 5, 8) pg 276(# 21, 25) pg 48 Dennison PreCalc (find VA, HA, SA, y int) pg 277 (# 29,31) ◦ pg 276 (# 7,12,32,35,41,42) Quiz on Graphs Tomorrow (No Graphing Calculators!) Tomorrow nights homework: pg 279 (# 1,4,9,10,11,13,16-19,24,25,27, 29,31,35)Sketch graph
P. 227 #1,5,7,15,16,18,23,25,34,41,50
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