Polynomial Degree and Finite Differences (Continued)

Home , Degree of a polynomial, Quadratic equation, Quadratic function

CONDENSED LESSON Polynomial Degree and 7.1 Finite Differences

In this lesson you will ● learn the terminology associated with polynomials ● use the finite differences method to determine the degree of a polynomial ● find a polynomial function that models a set of data A polynomial in one variable is any expression that can be written in the form

n n1 1 an x an1x · · · a 1x a 0 where x is a variable, the exponents are nonnegative integers, the coefficients are n n1 real numbers, and an 0. A function in the form f (x) anx an1x · · · 1 a 1 x a 0 is a polynomial function. The degree of a polynomial or polynomial function is the power of the term with the greatest exponent. If the degrees of the terms of a polynomial decrease from left to right, the polynomial is in general form. The polynomials below are in general form. 1st degree 2nd degree 3rd degree 4th degree 3x 7 x 2 2x 1.8 9x 3 4x 2 x 11 5x 4 A polynomial with one term, such as 5x 4, is called a monomial. A polynomial with two terms, such as 3x 7, is called a binomial. A polynomial with three terms, such as x 2 2x 1.8, is called a trinomial. Polynomials with more than three terms, such as 9x 3 4x 2 x 11, are usually just called polynomials. For linear functions, when the x-values are evenly spaced, the differences in the corresponding y-values are constant. This is not true for polynomial functions of higher degree. However, for 2nd-degree polynomials, the differences of the

differences, called the second differences and abbreviated D2, are constant. For 3rd-degree polynomials, the differences of the second differences, called the third

differences and abbreviated D3, are constant. This is illustrated in the tables on page 379 of your book. If you have a set of data with equally spaced x-values, you can find the lowest possible degree of a polynomial function that fits the data (if there is a polynomial function that fits the data) by analyzing the differences in y-values. This technique, called the finite differences method, is illustrated in the example in your book. Read that example carefully. Notice that the finite differences method determines only the degree of the polynomial. To find the exact equation for the polynomial function, you need to find the coefficients by solving a system of equations or using some other method.

In the example, the D2 values are equal. When you use experimental data, you may have to settle for differences that are nearly equal.

Investigation: Free Fall If you have a motion sensor, collect the (time, height) data as described in Step 1 in your book. If not, use these sample data. (The values in the last two columns are calculated in Step 2.) (continued)

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Complete Steps 2–6 in your book. The results Time (s) Height (m) given are based on the sample data. x y D1 D2

Step 2 The first and second differences, D1 and 0.00 2.000 D , are shown in the table at right. 0.012 2 0.05 1.988 0.025 For these data, we can stop with the second 0.037 0.10 1.951 0.024 differences because they are nearly constant. 0.061 0.15 1.890 0.025 0.086 0.20 1.804 0.024 0.110 0.25 1.694 0.025 0.135 0.30 1.559 0.024 0.159 0.35 1.400 0.025 0.184 0.40 1.216 0.024 0.208 0.45 1.008 Step 3 The three plots are shown below. (time, height ) (time 2, d 1) (time 3, d 2 )

Step 4 The graph of (time, height ) appears parabolic, suggesting that the correct model may be a 2nd-degree polynomial function. The graph of (time 2, d 1 ) shows that the first differences are not constant because they decrease in a linear fashion. The graph of (time 3, d 2 ) shows that the second differences are nearly constant, so the correct model should be a 2nd-degree polynomial function. Step 5 A 2nd-degree polynomial in the form y ax 2 bx c fits the data. Step 6 To write the system, choose three data points. For each point, write an equation by substituting the time and height values for x and y in the equation y ax 2 bx c. The following system is based on the values (0, 2), (0.2, 1.804), and (0.4, 1.216). c 2.000

0.04a 0.2b c 1.804

0.16a 0.4b c 1.216 One way to solve this system is by writing the matrix equation 0 0 1 a 2.000

0.04 0.2 1 1.804 b 0.16 0.4 1 c 1.216 and solving using an inverse matrix. The solution is a 4.9, b 0, and c 2, so an equation that fits the data is y 4.9x 2 2.

Read the remainder of the lesson in your book.

DDAA2CL_010_07.inddAA2CL_010_07.indd 9494 11/13/09/13/09 2:42:562:42:56 PPMM CONDENSED LESSON 7.2 Equivalent Quadratic Forms

In this lesson you will ● learn about the vertex form and factored form of a quadratic equation and the information each form reveals about the graph ● use the zero-product property to find the roots of a factored equation ● write a quadratic model for a data set in vertex, general, and factored form A 2nd-degree polynomial function is called a quadratic function. In Lesson 7.1, you learned that the general form of a quadratic function is y ax 2 bx c. In this lesson you will explore other forms of a quadratic function. You know that every quadratic function is a transformation of y x 2. When ____y k ____x h2 ____x h2 a quadratic function is written in the form b a or y b a k, you can tell that the vertex of the parabola is (h, k) and that the horizontal and vertical scale factors are a and b. Conversely, if you know the vertex of a parabola and you know (or can find) the scale factors, you can write its equation in one of these forms. ____x h2 The quadratic function y b a k can be rewritten in the form b b __ 2 __ y a 2( x h) k. The coefficient a 2 combines the two scale factors into one vertical scale factor. In the vertex form of a quadratic equation, y a(x h)2 k, this single scale factor is simply denoted a. From this form, you can identify the vertex, (h, k), and the vertical scale factor, a. If you know the vertex of a parabola and the vertical scale factor, you can write an equation in vertex form. Work through Example A carefully. The zero-product property states that for all real numbers a and b, if ab 0, then a 0, or b 0, or a 0 and b 0. For example, if 3x (x 7) 0, then 3x 0 or x 7 0. Therefore, x 0 or x 7. The solutions to an equation in the form f (x) 0 are called the roots of the equation, so 0 and 7 are the roots of 3x (x 7) 0. The x-intercepts of a function are also called the zeros of the function (because the corresponding y-values are 0). The function y 1.4(x 5.6)(x 3.1), given in Example B in your book, is said to be in factored form because it is written as the product of factors. The zeros of the function are the solutions of the equation 1.4(x 5.6)(x 3.1) 0. Example B shows how you can use the zero-product property to find the zeros of the function. In general, the factored form of a quadratic function is y a x r1 x r2 . From this form, you can identify the x-intercepts (or zeros), r1 and r2, and the vertical scale factor, a. Conversely, if you know the x-intercepts of a parabola and know (or can find) the vertical scale factor, then you can write the equation in factored form. Read Example C carefully.

Investigation: Rolling Along Read the Procedure Note and Steps 1–3 in your book. Make sure you can visualize how the experiment works. Use these sample data to complete Steps 4–8, and then compare your results to those below. (These data have been adjusted for the position of the starting line as described in Step 3.) (continued)

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Time (s) Distance from Time (s) Distance from Time (s) Distance from x line (m), y x line (m), y x line (m), y 0.2 0.357 2.2 3.570 4.2 3.309 0.4 0.355 2.4 3.841 4.4 2.938 0.6 0.357 2.6 4.048 4.6 2.510 0.8 0.184 2.8 4.188 4.8 2.028 1.0 0.546 3.0 4.256 5.0 1.493 1.2 1.220 3.2 4.257 5.2 0.897 1.4 1.821 3.4 4.193 5.4 0.261 1.6 2.357 3.6 4.062 5.6 0.399 1.8 2.825 3.8 3.871 5.8 0.426 2.0 3.231 4.0 3.619 6.0 0.419

Step 4 At right is a graph of the data. The data have a parabolic shape, so they can be modeled with a quadratic function. Ignoring the first and last few data points (when the can started and stopped), the second differences, D2, are almost constant, at around 0.06, which implies that a quadratic model is appropriate. Step 5 The coordinates of the vertex are (3.2, 4.257). Consider (5.2, 0.897) to be the image of (1, 1). The horizontal and vertical distances of (1, 1) from the vertex of y x 2 are both 1. The horizontal distance of (5.2, 0.897) from the vertex, (3.2, 4.257), is 2, and the vertical distance is 3.36. So, the horizontal and vertical scale factors are 2 and 3.36, respectively. This can be represented as the single vertical scale factor 3.36 ____ 22 0.84. Therefore, the vertex form of a model for the data is y 0.84(x 3.2)2 4.257. Step 6 Substituting the points (1, 0.546), (3, 4.256), and (5, 1.493) into the general form, y ax 2 bx c, gives the system a b c 0.546 9a 3b c 4.256

25a 5b c 1.493 The solution to this system is a 0.81, b 5.09, and c 3.74, so the general form of the equation is y 0.81x 2 5.09x 3.74. Step 7 The x-intercepts are about (0.9, 0) and (5.5, 0). The scale factor, found in Step 5, is 0.84. So the factored form of the equation is y 0.84(x 0.9)(x 5.5). Step 8 In general, you use the vertex form when you know either the vertex and the scale factor or the vertex and one other point you can use to find the scale factor. You use the general form when you know any three points. You use the factored form when you know the x-intercepts and at least one other point you can use to find the scale factor.

DDAA2CL_010_07.inddAA2CL_010_07.indd 9696 11/13/09/13/09 2:42:572:42:57 PPMM CONDENSED LESSON 7.3 Completing the Square

In this lesson you will ● use the method of completing the square to find the vertex of a parabola whose equation is given in general form ● solve problems involving projectile motion Many real-world problems involve finding the minimum or maximum value of a function. For quadratic functions, the maximum or minimum value occurs at the vertex. If you are given a quadratic equation in vertex form, you can easily find the coordinates of the parabola’s vertex. It is also fairly straightforward to find the vertex if the equation is in factored form. It gets more complicated if the equation is in general form. In this lesson you will learn a technique for converting a quadratic equation from general form to vertex form. Projectile motion—the rising or falling of objects under the influence of gravity—can be modeled by quadratic functions. The height of a projectile depends on the height from which it is thrown, the upward velocity with which it is thrown, and the effect of gravity pulling down on the object. The height can be modeled by the function 1 y __ gx 2 v x s 2 0 0 where x is the time in seconds, y is the height (in m or ft), g is the acceleration 2 2 due to gravity (either 9.8 m/s or 32 ft/s ), v0 is the initial upward velocity of the object (in either m/s or ft/s), and s0 is the initial height of the object (in m or ft). Read Example A in your book. It illustrates how to write a projectile motion equation when you know only the x-intercepts and how to use the x-intercepts to find the coordinates of the vertex.

Investigation: Complete the Square Complete the investigation in your book. When you are finished, compare your answers to those below. Step 1 a. x 2 5x 5x 25 x 2 10x 25 x 5 xx 2 5x b. (x 8) is the binomial expression being squared, and x 2 16x 64 is the perfect square trinomial. 55x 25

c. (x 12)2 x 12

2 d. a 2 2ab b 2 x x 12x 2 The first term of the trinomial, a , is the square of the first 12 12x 144 term of the binomial. The second term of the trinomial, 2ab, is twice the product of the binomial terms. The third term of the trinomial, b 2, is the square of the last term of the binomial. Step 2 a. You must add 9. b. x 2 6x x 2 6x 9 9 (x 3)2 9 2 2 c. Enter x 6x as f1(x) and (x 3) 9 as f 2(x), and verify that the table values or the graphs are the same for both expressions. (continued)

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Step 3 a. 9 b. x 2 6x 4 x 2 6x 9 9 4 (x 3)2 13 2 2 c. Enter x 6x 4 as f1(x) and (x 3) 13 as f2(x), and verify that the table values or the graphs are the same for both expressions. Step 4 a. Focus on x 2 14x. To complete a perfect square, you need to add 49. You need to subtract 49 to compensate. So

x 2 14x 3 x 2 14x 49 49 3 (x 7)2 46

2 _b2 __b 2 b. To make x bx a perfect square, you must add 2 , or 4 . You need to __b 2 subtract 4 to compensate. So b 2 b 2 b 2 b 2 x 2 bx 10 x 2 bx __ __ 10 x __ __ 10 4 4 2 4 Step 5 a. 2x 2 6x 1 2(x 2 3x) 1 Factor 2x 2 6x.

2 __9 __9 _9 2 x 3x 2 1 Complete the square. You add 2 и 4 , so you 4 4 и _9 must subtract 2 4 . 3 2 7 2 x __ Write in the form a (x h)2 k. 2 2

2 2 ___10 2 b. ax 10x 7 a x a x 7 Factor ax 10x.

2 ___10 ___25 ___25 и __25 a x x 2 a 2 7 Complete the square. You add a a 2 , so you a a a и __25 must subtract a a 2 . 2 __5 ___25 2 a x a 7 a Write in the form a (x h) k. __b __b 2 Step 6 The x-coordinate is 2a . Substitute 2a for x in y ax bx c to b 2 ___ find the y-coordinate, which is c 4a 2 .

Read Example B carefully. Based on your work in the investigation and Example B, you now know two ways to find the vertex, (h, k), of a quadratic function given in general form, y ax 2 bx c. 1. You can use the process of completing the square to rewrite the equation in vertex form, y a(x h)2 k. The vertex is (h, k). __b __b 2 2. You can use the formulas h 2a and k c 4a to calculate the coordinates of the vertex directly. You can use either method, but make sure you are comfortable with completing the square because it will come up in your later work. Example C applies what you learned in the investigation to solve a projectile motion problem. Try to solve the problem on your own before reading the solution.

DDAA2CL_010_07.inddAA2CL_010_07.indd 9898 11/13/09/13/09 2:42:582:42:58 PPMM CONDENSED LESSON 7.4 The Quadratic Formula

In this lesson you will ● learn how the quadratic formula is derived ● use the quadratic formula to solve projectile motion problems You can use a graph to approximate the x-intercepts of a quadratic function. If you can write the equation of the function in factored form, you can find the exact values of its x-intercepts. However, most quadratic equations cannot easily be converted to factored form. In this lesson you will learn a method that will allow you to find the exact x-intercepts of any quadratic function. Read Example A in your book carefully, and then read the example below.

EXAMPLE Find the x-intercepts of y 2x 2 7x 1.

Solution The x-intercepts are the solutions of 2x 2 7x 1 0. See if you can supply the reason for each step in the solution below. 2x 2 7x 1 0 2x 2 7x 1

7 2 x 2 __ x ? 1 ? 2 7 49 49 2 x 2 __ x ___ 1 ___ 2 16 8 7 2 41 2 x __ ___ 4 8 7 2 41 x __ ___ 4 16 ___ 7 41 x __ ______4 4 ______7 41 7 41 x __ _____ ______4 4 4 ______ ______7 41 ______7 41 The x-intercepts are x 4 3.351 and x 4 0.149.

The series of equations after Example A in your book shows how you can derive the quadratic formula by following the same steps used above. The quadratic formula, ______b b 2 4ac x ______2a

gives the general solution to a quadratic equation in the form ax 2 bx c 0. Follow along with the steps in the derivation, using a pencil and paper. To make

sure you understand the quadratic___ formula, use it___ to verify that the solutions of 2 ______7 41 ______7 41 2x 7x 1 0 are x 4 and x 4 . (continued)

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Investigation: How High Can You Go? Complete the investigation in your book, and then compare your answers to those below. Step 1 The equation is y 16x 2 88x 3, where y is the height in feet and x is the time in seconds. (If you answered this question incorrectly, review the discussion of projectile motion in Lesson 7.3.) Step 2 The equation is 24 16x 2 88x 3. Step 3 In ax 2 bx c 0 form, the equation is 16x 2 88x 21 0. For this equation, a 16, b 88, and c 21. Substituting these values into the quadratic formula gives ______2 88 88 4( 16)(21) 88 6400 88 80 x ______ ______ ______2(16) 32 32 ______88 80 ______88 80 So x 32 0.25 or x 32 5.25. The ball is 24 feet above the ground 0.25 second after it is hit (on the way up) and 5.25 seconds after it is hit (on the way down). Step 4 The y-coordinate of the vertex is 124. The ball reaches the maximum height only once. The ball reaches other heights once on the way up and once on the way down, but the maximum point is the height where the ball changes directions, so only one x-value corresponds to this y-value. Step 5 The equation is 124 16x 2 88x 3. In ax 2 bx c 0 form, the equation is 16x 2 88x 121 0. For this equation, a 16, b 88, and c 121. Substituting these values into the quadratic formula gives ______2 88 88 4(16)( 121) 88 0 88 x ______ ______ ____ 2.75 2(16) 32 32 The ball reaches a maximum height 2.75 seconds after it is hit. The fact that there is only one solution becomes apparent when you realize that the value under the square root sign is 0. Step 6 The equation is 200 16x 2 88x 3. In ax 2 bx c 0 form, the equation is 16x 2 88x 197 0. For this equation, a 16, b 88, and c 197. Substituting these values into the quadratic formula gives ______2 88 88 4(16)( 197) 88 4864 x ______ ______2(16) 32 The value under the square root sign is negative. Because the square root of a negative number is not a real number, the equation has no real-number solution.

Your work in the investigation shows that when the value under the square root sign, b 2 4ac, is 0, the equation ax 2 bx c 0 has only one solution, and when the value under the square root sign, b 2 4ac, is negative, the equation ax 2 bx c 0 has no real-number solutions. This means that if you are given a quadratic equation in the general form, you can use the value of b 2 4ac to determine whether the graph will have zero, one, or two x-intercepts. Example B in your book shows the importance of writing an equation in general form before you attempt to apply the quadratic formula. Read the example carefully.

DDAA2CL_010_07.inddAA2CL_010_07.indd 100100 11/13/09/13/09 2:42:592:42:59 PPMM CONDENSED LESSON 7.5 Complex Numbers

In this lesson you will ● learn that some polynomial equations have solutions that are complex numbers ● learn how to add, subtract, multiply, and divide complex numbers The graph of y x 2 x 2.5 has no x-intercepts. y

x –4 –2 2 4 –2 –4

If you use the quadratic formula to attempt to find the x-intercepts, you get ______2 1 1 4(1)(2.5) 1 9 x ______ ______2(1) 2 ______1 9 ______1 9 The numbers 2 and 2 are not real numbers because they involve the square root of a negative number. Numbers that include the real numbers as well as the square roots of negative numbers are called complex numbers. Defining the set of complex numbers makes it possible to solve equations such as x 2 x 2.5 0 and x 2 4 0, which have no solutions in the set of real numbers.

The square roots of negative numbers___ are expressed using an ___imaginary__ unit___ called i, defined by i 2 1 or i 1 . You can rewrite 9 as 9 и 1 , or 3i. Therefore, the two solutions to the quadratic equation above can be written ______1 3i ______1 3i _1 _3 _1 _3 as 2 and 2 , or 2 2 i and 2 2 i. These two solutions are a conjugate pair, meaning that one is in the form a + bi and the other is in the form a bi. The two numbers in a complex pair are complex conjugates. Roots of polynomial equations can be real numbers or nonreal complex numbers, or there may be some of each. However, as long as the polynomial has real-number coefficients, any nonreal roots will come in conjugate pairs, such as 3i and 3i or 6 5i and 6 5i. Your book defines a complex number___ as a number in the form a bi, where a and b are real numbers and i 1 . The number a is called the real part, and the number bi is called the imaginary part. The set of complex numbers includes all real numbers and all imaginary numbers. Look at the diagram on page 410 of your book, which shows the relationship between these numbers, and some other sets of numbers you may be familiar with, as well as examples of numbers in each set. Then read the example in your book, which shows how to solve the equation x 2 3 0. (continued)

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Investigation: Complex Arithmetic In this investigation you discover the rules for computing with complex numbers. Work through the investigation in your book before reading the answers below. Part 1: Addition and Subtraction Adding and subtracting complex numbers is similar to combining like terms. Use your calculator to add or subtract the numbers in Part 1a–d in your book. Then, make a conjecture about how to add complex numbers without a calculator. Below are the solutions and a possible conjecture. a. 5 i b. 5 3i c. 1 9i d. 3 i Possible conjecture: To add two complex numbers, add the real parts and add the imaginary parts. In symbols, (a bi ) (c di ) (a c) (b d )i. Part 2: Multiplication Multiplying the complex numbers a bi and c di is very similar to multiplying the binomials a bx and c dx. You just need to keep in mind that i 2 1. Multiply the complex numbers in Part 2a–d, and express the answers in the form a bi. The answers are below. a. (2 4i )(3 5i ) 2 и 3 2 и 5i 4i и 3 4i и 5i Expand as you would for a product of binomials.

6 10i 12i 20i 2 Multiply within each term.

6 2i 20i 2 Combine 10i and 12i.

6 2i 20(1) i 2 1

26 2i Combine 6 and 20. b. 16 3i c. 12 16i d. 8 16i Part 3: The Complex Conjugates Complete Part 3a–d, which involves finding either the sum or product of a complex number and its conjugate. The answers are below. a. 4 b. 14 c. 20 d. 32 Possible generalizations: The sum of a number and its conjugate is a real number: (a bi ) (a bi ) 2a. The product of a real number and its conjugate is a real number: (a bi )(a bi ) a 2 b 2. (continued)

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Part 4: Division To divide two complex numbers, write the division problem as a fraction, conjugate of denominator ______multiply by conjugate of denominator (to change the denominator to a real number), and then write the result in the form a bi. Divide the numbers in Part 4a–d. Here are the answers. ______7 2i ______7 2i _____1 i conjugate of denominator a. и Multiply by ______. 1 i 1 i 1 i conjugate of denominator 5 9i ______Multiply. The denominator becomes a real number. 2

2.5 4.5i Divide. b. 0.5 2i c. 0.22 0.04i d. 0.6 0.8i

Complex numbers can be graphed on a complex plane, where the horizontal axis is the real axis and the vertical axis is the imaginary axis. The number a bi is represented by the point with coordinates (a, b). The numbers 3 4i and 4 i are graphed below.

Imaginary axis 5 3 4i 4 i Real axis –5 5

–5

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In this lesson you will ● learn about cubic functions ● use the x-intercepts of a polynomial function to help you write the function in factored form The polynomial equations y x 2 6x 9 and y (x 3)(x 3) are equivalent. The first is in general form, and the second is in factored form. Writing a polynomial equation in factored form is useful for finding the x-intercepts, or zeros, of the function. In this lesson you will learn some techniques for writing higher-degree polynomials in factored form. A 3rd-degree polynomial function is called a cubic function. At right y is a graph of the cubic function y 4x 3 16x 2 9x 36. 50 The x-intercepts of the function are 4, 1.5, and 1.5, so its factored (0, 36) equation must be in the form y a (x 4)(x 1.5)(x 1.5). To find the (–4, 0) (1.5, 0) value of a, you can substitute the coordinates of another point on the curve. x The y-intercept is (0, 36). Substituting this point into the equation gives –5 5 36 a (4)(1.5)(1.5). So, a 4, and the factored form of the equation is (–1.5, 0) y 4(x 4)(x 1.5)(x 1.5) –50 Read the text before Example A in your book and then work through Example A.

Investigation: The Box Factory You can make a box from a 16-by-20-unit sheet of paper by xx cutting squares of side length x from the corners and folding x x the sides up.

Follow the Procedure Note in your book to construct boxes for 16 several different integer values of x. Record the dimensions and volume of each box. (If you don’t want to construct the boxes, try to picture them in your mind.) Complete the investigation, x x and then compare your results to those below. xx 20 Step 1 Here are the results for integer x-values from 1 to 6.

Volume x Length Width Height y 1 18 14 1 252 2 16 12 2 384 3 14 10 3 420 4 12 8 4 384 5 10 6 5 300 6 8 4 6 192

Step 2 The dimensions of the boxes are 20 2x, 16 2x, and x. Therefore, the volume function is y (20 2x)(16 2x)(x). (continued)

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Step 3 The data points lie on the graph of the function.

Step 4 If you were to expand (20 2x)(16 2x)(x), the result would be a polynomial, and the highest power of x would be 3. Also, the graph looks like a cubic function. Therefore, the function is a 3rd-degree polynomial function. Step 5 The x-intercepts of the graph are 0, 8, and 10, so the function is y x (x 8)(x 10). Step 6 The graphs have the same x-intercepts and general shape but different vertical scale factors. A vertical scale factor of 4 makes them equivalent: y 4x (x 8)(x 10). Step 7 If x 0, there are no sides to fold up, so a box cannot be formed. For x 8, 8-unit-wide strips would be cut off the sides of the sheet. Folding up the “sides” would mean folding the remaining strip in half, which would not form a box. 88

Cut Cut 8 8 off off 16 Cut Cut 8 8 off off

88 20 A value of x 10 is impossible because it is more than half the length of the shorter side of the sheet. Only a domain of 0 x 8 makes sense in this situation. By zooming and tracing to find the coordinates of the high point of the graph, you can find that the x-value of about 2.94 maximizes the volume.

Work through Example B in your book, which asks you to determine the factored form of a polynomial function by using the x-intercepts of the graph. This method works well when the zeros of a function are integer values. Unfortunately, this is not always the case. Sometimes the zeros of a polynomial are not “nice” rational or integer values, and sometimes they are not even real numbers. With quadratic functions, if you cannot find the zeros by factoring or making a

graph, you can always use the quadratic formula. Once you know the zeros, r1 and r2, you can write the polynomial in the form y a x r1 x r2 . Read the remainder of the lesson in your book, and then read the example on the next page. (continued)

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EXAMPLE Write the equation of the quadratic function below in factored form. y 2 x –2 2 4 6 8 –2 (2, –2) –4 –6 –8

Solution The factored equation is in the form y a x r1 x r2 , where r1 and r2 are the zeros. From the graph, you can see that the only real-number zero is 3. If the other zero were a nonreal number, then its conjugate would also be a zero. This would mean there are three zeros, which is not possible. So 3 must be a “double zero.” This means that the function is in the form y a (x 3)(x 3), or y a (x 3)2. To find the value of a, substitute (2, 2): 2 a (1)2, so a 2. The factored form of the function is y 2(x 3)2.

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In this lesson you will ● describe the extreme values and end behavior of polynomial functions ● solve a problem that involves maximizing a polynomial function ● write equations for polynomial functions with given intercepts Polynomials with degree 3 or higher are often referred to as higher-degree y polynomials. At right is the graph of the polynomial y x (x 3)2, or y x 3 6x 2 9x. The zero-product property tells you that the zeros are x 0 and x 3. These are the values of x for which y 0. The x-intercepts of the graph confirm this. 2 The graph has other key features in addition to the x-intercepts. For example, x –2 2 4 6 the point (1, 4) is called a local minimum because it is lower than the other –2 points near it. The point (3, 0) is called a local maximum because it is higher –4 than the other points near it. You can also describe the end behavior of the (1, –4) graph—that is, what happens to the graph as x takes on extreme values in the –6 positive and negative directions. On this graph, look at values of x greater than 4. As x increases, y decreases. Now look at negative values of x. As x decreases, y increases. The introduction to Lesson 7.7 in your book gives another example of a 3rd-degree polynomial and its graph. The graph of a polynomial function with real coefficients has a y-intercept, possibly one or more x-intercepts, and other features such as local maximums or minimums and end behavior. The maximums and minimums are called extreme values.

Investigation: The Largest Triangle Start with a 21.5-by-28 cm sheet of paper. Orient the paper so that the long side is horizontal. Fold the upper left corner so that it touches some point on the bottom edge. Find the area, in cm2, of the triangle AA formed in the lower left corner. x x What distance, x, along the bottom edge of the paper produces the triangle with greatest area? To answer this question, first find areas for different values of x. Then find a formula for the area of the triangle in terms of x. Try to do this on your own before reading on. Your answers will vary depending on the dimensions of your paper, but the logic used should still apply. h 21.5 h Here is one way to find the formula: Let h be the height of the triangle. Then the hypotenuse has length 21.5 h. (Why?) x (continued)

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Use the Pythagorean Theorem to help you write h in terms of x . 462.25 x 2 x 2 h 2 (21.5 h)2, so h ______43 Now, you can write a formula for the area, y. 1 462.25 x 2 y __ x ______ 2 43 At right is a graph of the area function and some sample data points. If you trace the graph, you’ll find that the maximum point is about (12.4, 44.5). Therefore, the value of x that gives the greatest area is about 12.4 cm. The maximum area is about 44.5 cm2.

Example A in your book shows you how to find the equation for a polynomial with given x-intercepts and nonzero y-intercept. Read this example carefully. To test your understanding, find a polynomial function with x-intercepts 6, 2, and 1, and y-intercept 60. (One answer is y 5(x 6)(x 2)(x 1).) Graphs A–D on page 425 of your book show some possible shapes for the graph of a 3rd-degree polynomial function. Graph A is the graph of the parent function y x 3. Like other parent functions you have studied, the graph can be translated, dilated, and reflected. Example B in your book shows you how to find a polynomial function with given zeros when some of the zeros are complex. The key to finding the solution is to recall that nonreal complex zeros come in conjugate pairs. Read that example carefully, and then read the example below.

EXAMPLE Find a 4th-degree polynomial function with real coefficients and zeros x 2, x 3, and x 1 i.

Solution Nonreal complex zeros of polynomials with real coefficients occur in conjugate pairs, so x 1 i must also be a zero. So one possible function, in factored form, is y (x 2)(x 3)[x (1 i)][x (1 i)] Multiply the factors to get a polynomial in general form. y (x 2)(x 3)[x (1 i)][x (1 i)] x 2 x 6x 2 (1 i )x (1 i )x (1 i )(1 i ) x 2 x 6x 2 x ix x i x 2 x 2 x 6x 2 2x 2 x 4 2x 3 2x 2 x 3 2x 2 2x 6x 2 12x 12 x 4 3x 3 2x 2 10x 12 Check the solution by making a graph. (You will see only the real zeros.)

Note that an nth-degree polynomial function always has n zeros (counting multiple roots). However, because some of the zeros may be nonreal numbers, the function may not have n x-intercepts.

DDAA2CL_010_07.inddAA2CL_010_07.indd 110110 11/13/09/13/09 2:43:022:43:02 PPMM CONDENSED LESSON 7.8 More About Finding Solutions

In this lesson you will ● use long division to find the roots of a higher-degree polynomial ● use the Rational Root Theorem to find all the possible rational roots of a polynomial ● use synthetic division to divide a polynomial by a linear factor You can find the zeros of a quadratic function by factoring or by using the quadratic formula. You can sometimes use a graph to find the zeros of higher- degree polynomials, but this method may give only an approximation of real zeros and won’t work at all to find nonreal zeros. In this lesson you will learn a method for finding the exact zeros, both real and nonreal, of many higher-degree polynomials. Example A in your book shows that if you know some of the zeros of a polynomial function, you can sometimes use long division to find the other roots. Follow along with this example, using a pencil and paper. Make sure you understand each step. To confirm that a value is a zero of a polynomial function, you substitute it into the equation to confirm that the function value is zero. This process uses the Factor Theorem, which states that (x r) is a factor of a polynomial function P (x) if and only if P (r) 0. When you divide polynomials, be sure to write both the divisor and the dividend so that the terms are in order of decreasing degree. If a degree is missing, insert a term with coefficient 0 as a placeholder. For example, to divide x 4 13x 2 36 by x 2 9, rewrite x 4 13x 2 36 as x 4 0x 3 13x 2 0x 36 and rewrite x 2 9 as x 2 0x 9. The division problem below shows that x 4 13x 2 36 x 2 9x 2 4.

2 ______x 4 x 2 0x 9 ) x 4 0x 3 13x 2 0x 36 x 4 0x 3 9x 2 4x 2 0x 36 4x 2 0x 36 0

In Example A, you found some of the zeros by looking at the graph. If the x-intercepts of a graph are not integers, identifying the zeros can be difficult. The Rational Root Theorem tells you which rational numbers might be zeros. It states that if the polynomial equation P (x) 0 has rational roots, then they are _p of the form q , where p is a factor of the lowest-degree term and q is a factor of the leading coefficient. Note that this theorem helps you find only rational roots. Example B shows how the theorem is used. Work through that example, and then read the example on the next page. (continued)

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EXAMPLE Find the roots of 7x 3 3x 2 56x 24 0.

Solution The graph of the function y = 7x 3 3x 2 56x 24 appears at right. None of the x-intercepts are integers. The Rational Root Theorem tells you that any rational root will be a factor of 24 divided by a factor of 7. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The factors of 7 are 1 and 7. You know there are no _1 _2 _3 integer roots, so you need to consider only 7 , 7 , 7 , _4 _6 _8 __12 __24 7 , 7 , 7 , 7 , and 7 . The graph indicates that one of the roots is between 3 and 2. None of these possibilities _1 are in that interval. Another root is a little less than 2 . This _3 _3 could be 7 . Try substituting 7 into the polynomial. 3 3 3 2 3 27 27 7 __ 3 __ 56 __ 24 ___ ___ 24 24 0 7 7 7 49 49 _3 _3 So 7 is a root, which means x 7 is a factor. Use long division to divide out this factor.

2 ______7x 56 _3 ) 3 2 x 7 7x 3x 56x 24 7x 3 3x 2 56x 24 56x 24 0 3 So 7x 3 3x 2 56x 24 0 is equivalent to x _ 7x 2 56 0. To find the 7 __ __ other roots, solve 7x 2 56 0. The solutions are x 8 , or 2 2 . So the __ __ _3 roots are x 7 , x 2 2 , and x 2 2 .

Synthetic division is a shortcut method for dividing a polynomial by a linear factor. Read the remainder of the lesson in your book to see how to use synthetic 7x 3 3x 2 56x 24 division. Below is an example using synthetic division to find ______. __3 x 7 Note that in the example above you found this same quotient using long division. Known zero Coefficients of 7x3 3x 2 56x 24 _3 7 7 3 56 24 1 Bring 3 Add 5 Add7 Add down _3 и 30_3 и _3 и Ϫ 24 2 7 7 4 7 0 6 7 56 7056 0

______7x 3 3x 2 56x 24 2 3 2 The result shows that 3 7x 56, so 7x 3x 56x 24 x __ _3 2 7 x 7 7x 56 .

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