Understanding Polynomials and Graphs through Real-World Analogies
Explore the relationship between mountain ranges and polynomials, and learn how to apply the Intermediate Value Theorem to find zeros of polynomial functions. This guide covers concepts like the Interval Value Theorem, sketching graphs of higher-degree polynomials, and constructing tables to analyze a function's behavior. Discover practical insights using visuals to deepen your understanding of polynomial functions and their graphs.
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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<b, then f takes on every value between f(a) and f(b) in the interval [a, b]. Facts about IVT: f is continuous from f(a) to f(b) If f(a) and f(b) have opposite signs (1 positive and 1 negative), there is at least one number c between a and b such that f(c)=0.
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)= x +2x -6x +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 ( ) 1 , Interval , 4 ( ) ) 4 , 1 ( Sign of (x-1) - - + + Sign of (x-4) - - - - + Sign of f(x) + - - + Position of graph Above x- axis Below x- axis Above x- axis ) 1 , and ) 4 , 1 ( ( This means that f(x)>0 if x is in This means that f(x)<0 if x is in . , 4 ( )
Sketching the graph of a polynomial function of degree 3 Ex Step 1: Find Zeros Group terms Factor out x and -4 Factor out (x+1) Difference of squares Therefore, the zeros are -2, 2, and -1 = + 3 2 ( ) 4 4 f x x x x = + + 3 2 ( ) ( ) ( 4 ) 4 f x x x x = ) ) 1 + )( ) 1 + ) 1 + 2 ( ) x ( x ( 4 x f x ( x x x = 2 ( 4 f = + ) 1 + ( ) ( 2 )( 2 )( f x x x x
Step 2: Create a Table Interval ) 2 , 2 ) 1 , 2 ( ) 2 , 1 ( ( , ( ) Sign of x+2 Sign of x+1 Sign of x-2 Sign of f(x) Position of graph
f(x)>0 if x is in f(x)<0 if x is in My version of graph (there are multiple ways to do this graph) y , 2 ) 1 , ) 2 , 1 ( ( , 2 ( ) 2 ) ( x
Ex Find the intervals f(x) is below x-axis when x is in f(x) is above x-axis when x is in Sketch graph , 3 ) 3 ), 1 ( , ) 2 , 0 ( ), 0 , 1 and ( ( , 2 ( ) and y 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 6 . 4 ) ( = x x f + 3 2 . 5 72 . 0 656 x x 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
P. 227 #1,5,7,15,16,18,23,25,34,41,50 If you need help, check with the person beside you.
We will first divide 2172 by 3 to find the quotient and remainder.
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 . 4 2 x 81 , 9 + x , 9 , 3 + 2 and 3 x x x Notice that is not divisible by . 4 + x + 2 81 3 1 x x However, we can use Polynomial Long Division to find a quotient and a remainder.
Step 1: terms are missing, use a zero to fill in the missing term (this will help with the spacing). Step 1: Make sure the polynomial is written in descending order. If any Step 2: symbol by the term with the highest power outside the division symbol. Step 2: Divide the term with the highest power inside the division Step 3: by the polynomial in front of the division symbol. Step 3: Multiply (or distribute) the answer obtained in the previous step Step 4: Step 4: Subtract and bring down the next term. Step 5: down. Step 5: Repeat Steps 2, 3, and 4 until there are no more terms to bring Step 6: step is the remainder and must be written as a fraction in the final answer. Step 6: Write the final answer. The term remaining after the last subtract http://www.youtube.com/watch?v=smsKMWf8ZCs
Divide by . + + + 3 2 2 + x 2 7 2 9 x x x 3 + + + 2 2 2 2 9 x x x x + + 3 2 + + + + 2 3 2 7 3 2 x x 2 3 2 7 2 9 x x x x + 3 2 2 3 x x + 2 4 2 x x + 2 4 6 x x + 4 9 x 4 6 x 15 + + + 3 2 2 7 2 9 15 x x x x = + + 2 Solution : 2 2 x x + + 2 3 2 3 x
Divide by . + 3 2 3 4 12 x x x + x 2 6 x + 2 x + + 2 3 2 + + 2 3 2 6 3 4 12 x x x x x 6 3 4 12 x x x x x + 3 2 6 + x x x 2 2 2 12 x x + 2 2 2 12 x x 0 + x 3 2 3 x 4 12 x x x = + Solution : 2 x + 2 6
If a polynomial f(x) is divided by x-c, the the remainder is f(c). Example: the use 5, 3 ) ( If + + = x x x x f 3 2 remainder theorem to find f(-2). + + + 3 2 2 3 5 x x x x = Therefore, (-2) -17. f
A polynomial f(x) has a factor x-c if and only if f(c)=0. Example: = + + 3 2 Show that x - a is 2 factor of ( ) 4 3 2. f x x x x
Find a polynomial f(x) of degree 3 that has zeros 3, -1, and 1.
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.
4 3 + + Let' compare s dividing x x 2 x 2 by x using 1 both long division Long Division synthetic and Long Division division. Synthetic Division Synthetic Division + x 3 2 x 2 + x 2 x 4 + 1 1 1 0 2 2 + 4 3 2 x 1 x x 0 x 2 2 1 2 2 4 4 3 x x + 3 2 1 2 2 4 6 2 x 0 x ( ) 3 2 2 x 2 x 3 x 2 2 + x 2 2 x 2 x 2 4 x ( ) + 2 2 x 2 x + 4 x 2 ( ) 4 x 4 6
Use synthetic division to find the quotient q(x) and remainder r ) ( if = x f + + 4 3 2 4 3 divided is 5, by . 3 x x x x
Using synthetic division to find zeros. What must we show for a value to be a zero of f(x)? Think Factor Theorem = + + 3 2 Show that - a is 11 zero of ( ) 8 29 44. f x x x x
Use synthetic division to find f(3) if 3 ) ( = x x f + + 4 3 2 3 15 x x x
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).
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 )( ( ) ( 1 x c x a x f = , ,..., c c nc 1 2 )...( ), c x nc 2 Where a is the leading coefficient of f(x). Each number is a zero of f(x). kc
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. 1 ) ( = x x f ) 3 ) 1 + 3 2 ( 2 )( ( x x 25
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? = ) 3 ) 1 + + 3 2 5 ( ) ( 3 2 )( ( ( ) 2 f x x x x x
If f(x) is a polynomial with real coefficients and a nonzero constant term, then: 1. The number of positive is equal to the number of variations of sign in f(x) or is less than that number by an even integer. positive real zeros of f(x) either 2. The number of negative is equal to the number of variations of sign in f(-x) or is less than that number by an even integer. negative real zeros of f(x) either
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 2 3 ) ( + = x x x f + 5 4 3 3 6 5 x x 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 = + + 4 3 2 ( ) 2 3 6 f x x x x x
