MATH 4 part 2

What have you learnedA. Write an equation to show the functional relationship between the twoquantities involved in the problems using the indicated variables. 1. The cost ( c ) of n cavans of rice at P1200 a cavan. 2. The time ( t ) required to make a trip of 60 km. if a car goes r km. per hour. 3. The surface area of a sphere S equals 4π times the square of the radius. 4. A number decreased by 12 is 7. 5. The sum of two consecutive integers is 49. B. Solve the following problems. 6. The amount of flour used in baking bread is directly proportional to the number of loaves of bread. If 6 cups of flour makes 2 loaves of bread, how many cups of flour would be used to make 5 loaves of bread? 7. A jet plane and a prop plane leave the same airport at the same time and travel in opposite directions. The jet travels at 960 km/hr and the prop plane travels at 560 km/h. In how many hours will they be 4560 km apart? 8. The sum of three consecutive even integers is 50 more than the third integer. find the integers. 9. To increase the sugar concentration of a solution from 15% to 40%, how many kilos of pure sugar must be added to 80 kilos of the 15% solution? 10. David is three years older than Jim. Tom is 5 years younger than Jim. The sum of their ages is 34 years. Find their ages. 23

Answer KeyHow much do you know1. A = s2 2. P = 2l + 2w 3. 12x + 7 = 3x – 44. C = 2πr7. 3 5. d = rt 6. 246 km.9. 13, Jean’s age 8. P120 is Gloria’s allowance 18, Kim’s age 10. 4, 5, 6Try this outLesson 11a. m 01 2 3 4 5 6 C(m) 2000 8000 14000 20000 26000 32000 38000 7 8 9 10 44000 50000 56000 62000b. 44,000 c. 8 km2. b. 8 cellphones c. P20,000.00 a. P3750 e. I = 5,000 + 150n d. 67 cellphones3. P11,9004. aD 1 23F 2600 3200 3,800b. P5600 3. 13 daysLesson 2 2. 0.84 kg 3. P1120.001. 22 1 liters 5. 10.5 cubic meters 24. 30 km 24

Lesson 3 2. 8x = 56 3. x + (x+1) +(x +2) = 99A.1. x – 12 = 8 5. 3(x + 3) = -18 4. m + 7, m + 14B. 2. x = 3 3. 191. 51 , 53, 55 5. -12, -10, -84. x = 10Lesson 4 b. m – 2A. d. (m -5) + (m – 7) = 36a. m – 5c. m – 7B.1. 5, age of Lita ; 7, age of Susie2. 10 yrs. old3. Rico – 4 , Joyce – 84. 10 yrs.5. 10 yrs. oldLesson 51. 20 - 25¢ & 10 - 50¢2. 17 – P1.00 & 13 – P5.003. 455 – P50.00 & 230 – P75.004. 9 - 50¢ & 1 - 25¢5. 3 bottles of shampoo & 9 soap barsLesson 6 2. 87.5 l 3. 540 gA.1. 37.5 mlB.1. 13.33 kilos of water 2. 4 liters 3. 7.5 kl – amount of peanuts 12.5 kl – amount of green peas4. 12.75 kg of P29/kl cheese 7.5 kg of P31/kl cheese 25

Lesson 7 1. 22 km/hr 2. 240 miles/hr 3. 24 minutes 4. 3 1 hrs 5What have you learned 1. C = P1200n 2. t = 60km r 3. S = 4πr2 4. n – 12 = 7 5. x + (x +1) = 49 6. 15 cups of flour 7. 3 hrs 8. 24, 26, 28 9. 33 1 kilos of pure sugar 3 10. Tom’ age – 7 yrs old Jim’s age – 12 yrs. old Davis’s age – 15 yrs. old 26

Module 3 Polynomial Functions What this module is about This module is about graphs of polynomial functions of degree greaterthan two. The graph of a first degree-polynomial is a line. The graph of a second-degree polynomial is a parabola. The graph of a third degree- polynomialtypically has both a minimum point and a maximum point. The number ofmaximum and minimum points is at most one less than the degree of thepolynomial. The graph of a polynomial function of degree n has n – 1 turningpoints. The lessons were presented in a very simple way so it will be easy for youto understand and be able to do the graphs of the polynomial functions of degreegreater than two without difficulty. What you are expected to learn This module is designed for you to: 1. identify the function represented by a graph. 2. draw the graphs of polynomial functions of degree greater than 2. 3. identify the degree of a polynomial function, its zeros, x-intercepts and turning points based on its equation and graph.

How much do you knowA. Identify whether the function represented by the graph below is linear, quadratic or polynomial.1. 2. Y YX X3. 4. Y Y X X5. Y X 2

B. From the graph and equation, find the following: degree, number of zeros, number of x-intercept and the number of turning points.1. Y Degree: ____________________ X No. of zeros: ________________ No. of x-intercepts: ___________ No. of turning points: __________2. Y Degree: _____________________ No. of zeros: _________________X Y No. of x- intercepts: _________ __ 3. No. of turning points: __________ X Degree: ______________________ No. of zeros: __________________ No. of x-intercepts: _____________ No. of turning points: ___________ 3

4. Y Degree: _________________ X No. of zeros: _____________ No. of x-intercepts: ________ No. of turning points: ______ Y Degree: ______________5. No. of zeros: __________ No. of x-intercepts: _____ X No. of turning points: ____C. Draw the graph of the polynomial function f(x) = (x + 3)(2x + 1)(x – 1)(x – 3). Determine the following after drawing the graph: a. behavior of the graph b. zeros and the y-intercept of the function. c. interval where G(x) ≥ 0. What you will do Lesson 1 Graphs of linear and Quadratic Functions You have learned that the graph of a first-degree polynomial is a line and asecond-degree polynomial is a parabola. At this point, recall the very importantsteps that are undertaken in the construction of a graph. The first step youusually take is to prepare a table of values for x and y that satisfy the given 4

equation. These ordered pairs are points that lie on the curve described by theequation. You then plot the points described by the ordered pairs in the table.Examples:1. Graph the line determined by the equation 4x - y = -2. The values of x and y is seen in table 1.1 while the graph is seen in Figure 1.1.Table 1.1 Figure 1.1 Y XY -2 -6 02 16 2 10 X Note that the values of x and y are obtained by direct substitution. Seethese computations.if x = -2, then y = 4x + 2 if x = 0, then y = 4x + 2 = 4(-2 ) + 2 = 4( 0 ) + 2 = -8 + 2 =2 = -6if x = 1, then y = 4x + 2 if x = 2, then y = 4x + 2 = 4( 1 ) + 2 = 4( 2 ) + 2 =4+2 = 8 + 10 =6 = 10 You have seen that the graph of a linear function is a slanting continuousline. A linear function is of the form f(x) = mx + b. The variable m is the slope ofthe line, m ≠ 0, and b is the y-intercept. The zero of the linear function is thevalue of the independent variable that makes the value of the function 0. It is the 5

x-intercept of the linear function. If the function is y = mx + b, the zero of y is thevalue of x such that mx + b = 0. The function is increasing when m 〉 0, anddecreasing when m 〈 0. You have learned that the domain of a linear function isthe set of real numbers and the range is also the set of real numbers. In the function y = 4x + 2, the degree is one, it has one x-intercept andone zero. It has no turning point since the graph is a slanting line. Another function that you have learned is the quadratic function whosegraph is a parabola that opens upward when a 〉 0 and downward when a 〈 0. Aquadratic function is a function that can be described by an equation of the formf(x) = ax2 + bx + c = 0, where a ≠ 0. The vertex (h, k) of the parabola is theturning point of the graph of a quadratic function.2. Draw the graph of the quadratic function y = x2 – 4x + 4.Solution: Find the vertex of the function.a = 1 b = -4 c = 4h = −b = 2 k = 4ac − b2 = 0 The vertex is (2, 0). 2a 4a Table 1.2X -2 -1 0 1 2 3 4 5 6y 16 9 4 1 0 1 4 9 16 Figure 1.2 YX 6

Try this out A. Identify at least three points of the given linear function. Then use these points to draw the graph of the function. 1. y = 5x – 8 2. y = -2x + 3 3. 2x + y – 3 = 0 B. Determine the vertex, find some representative points then draw the graph 4. y = x2 – 7 5. y = x2 - 6x + 9 Lesson 2A Graphing Polynomial Functions In graphing a polynomial function, the technique of finding and plotting asmany point as possible will be helpful. But there are theorems concerning roots ofpolynomial equations that will be of great help to obtain the sketch of the graph ofthe polynomial function. The graph of a third-degree polynomials or a cubic polynomial, has bothmaximum and minimum points. The number of maximum and minimum points is atmost one less than the degree of the polynomial. The graph of a polynomialfunction of degree n has n – 1 turning point. When you graph higher degree polynomial functions, the rational zerotheorem and Rene Descartes’ rule can help you find any integral zeros. If there aremany possibilities to try, it is also useful to know the upper bound and lower boundfor the zeros. An upper bound for the real zeros of a polynomial function is anumber greater than or equal to the greatest real zero of the function. Similarly, alower bound is a number less than or equal to the least real zero of the function. 7

Upper and Lower Bound Theorem let a polynomial function be divided by x – c. • If c 〉 0 and all the coefficients in the quotient and remainder are nonnegative, then c is an upper bound of the zeros. • If c 〈 0 and the coefficients in the quotient and remainder alternate in sign, then c is a lower bound of the zeros.Example: Find the upper and lower bounds of the zeros of P(x) = 2x4 + 5x3 - 3x2 - 9x + 1. Solution: It is important to emphasize that c is not restricted to integers. It can be any real number. however, integers are used in illustrative examples for convenience in the computation. Test c = 1 for upper bound 2 5 -3 -9 1 1 2 7 4 -5 2 7 4 -5 -4 The third entries are not all positive. So, 1 is not an upper bound. Test c = 2 2 5 -3 -9 1 2 4 18 30 42 2 9 15 21 43 The third entries are all positive, so 2 is an upper bound. Test c = 3 2 5 -3 -9 1 3 6 33 90 243 2 11 30 81 244 8

The numbers 3, 4, 5, … are upper bounds. The number 2 is said to be thesmallest integral upper bound. This means that there is no zero of P(x) largerthan 2. Test c = -3 2 5 -3 -9 1 -3 -6 3 0 27 2 -1 0 -9 28 The third entries are alternating in signs, so -3 is a lower bound. If the test for c = -4, -5, -6, … are made, the third entries are alternating.The numbers -4, -5, -6, … are lower bounds. The number -3 is said to be thelargest lower bound. This means that there is no zero of P(x) smaller that -3.Descartes’ Rule of Signs Another information that may be helpful in graphing polynomial functionsis the knowledge of the maximum number of positive and negative roots. Thisinformation can be provided by the application of Descartes’ Rule of signs. The rule makes use of the number of variations in sign of the coefficientsof the polynomial from left to right. The terms must be arranged in descendingpowers of x. A polynomial is said to have a variation in sign if two consecutiveterms have opposite signs.Descartes’ Rule of Signs Let p(x) = 0 be a polynomial equation with real coefficients, the leadingcoefficient an 〉 0, and is arranged with descending powers of x. • The number of positive roots of p(x) = 0 is either equal to the number of variations of signs in p(x), or is less than that number by an even counting number. • The number of negative roots of p(x) = 0 is either equal to the number of variations in signs in p(-x), or is less than that number by an even counting number. 9

Example: Determine all possible combinations of number of positive and negativeroots of the given polynomial function.1. f(x) = 2x4 + 5x3 -2x2 - 4x + 5f(x): + + -- + 2 variationsf(-x): + - -+ + 2 variationsNo. of positive roots No. of negative roots Total number of roots 2 2 4 f(x) = 0 has either 2 positive roots and 2 negative roots.2. f(x) = x5 - 5x4 - 3x3 + 15x2 - 4x + 20 f(x): + - - + - + 4 variationsf(-x): - - + ++ + 1 variationNo. of positive roots No. of negative roots Total number of roots 4 1 5 f ( x ) = 0 has 4 positive and 1 negative roots.Graph of Odd-degree Polynomials The extreme left and right parts of the graph of p( x ) = an xn + an-1 xn-1 +… + a2 x2 + a1 x + a0 are: 1. increasing; if n is odd and an 〉 0. 2. decreasing; if n is odd an 〈 0. The domain and range are the set of real numbers. 10

f(x) Y Y g(x) XXf(x) is a polynomial where n = 3 and an 〉 0g(x) is a polynomial where n = 5 and an 〈 0Graph of Even-degree PolynomialsThe graph of p( x ) = an xn + an-1 xn-1 + … + a2 x2 + a1 x + a0 has1. decreasing extreme left and increasing extreme right parts, n is even and an 〉 0.2. increasing extreme left and decreasing extreme right parts, n is even and an 〈 0. The domain is the set of real numbers and the range is the set ofnonnegaive numbers.F(x) Y G(x) YXXF(x) is a polynomial where n = 2 and an 〉 0.G(x) is a polynomial where n = 4 and an 〈 0. 11

Try this outA. Find the upper and lower bounds of the zeros of each polynomial function. 1. f ( x ) = x3 + 4x2 + 8x + 5 2. p ( x ) = 6x4 + x3 - 56x2 - 9x + 18 3. q ( x ) = x5 - 4x4 - 9x3 + 3x2 + 16x - 4B. Describe the graph of each polynomial based on its degree and leading coefficients. 4. f ( x ) = x3 + 4x2 + 8x + 5 5. p ( x ) = 6x4 + x3 - 56x2 - 9x + 18 6. q ( x ) = x5 - 4x4 - 9x3 + 3x2 + 16x - 4C. Use the Descartes’ Rule of Signs to determine the possible combinations of roots for each equation. 7. f ( x ) = x3 + 4x2 + 8x + 5 8. p ( x ) = 6x4 + x3 - 56x2 - 9x + 18 9. q ( x ) = x5 - 4x4 - 9x3 + 3x2 + 16x - 4 Lesson 2B Graphing Polynomial Functions To graph polynomial function of degree greater than two, evaluate thefunction to determine ordered pairs, then plot the points with this pairs ascoordinates and connect the points to form a smooth curve. The x-coordinates ofthe points where the graph meets the x-axis are the zeros of the function. Theinformation you can get from the given equation of a polynomial function are thedegree and the number of zeros. From the graph, you can determine the numberof x-intercepts, y-intercept and number of turning points. The theorems discussed 12

in the previous lesson concerning the roots of polynomial equations will be ofgreat help to obtain the sketch of the graph of the polynomial function.Examples:1. G( x ) = x3 - x2 - x + 1a. Determine the behavior of the graphb. Get the zeros and the y-intercept of the function.c. Sketch the graphd. Determine the interval where G( x ) ≥ 0.Solution:a. The degree of the function is 3 which is an odd, it is an odd-powered function behaving like a line that starts to the right. Since, the leading coefficient is positive. It is increasing.b. Using synthetic division1 -1 -1 1 1 1 0 -1 10 -1 0 x2 - 1 = 0 (x + 1)(x – 1) = 0x+1 = 0 x–1 = 0 x = -1 x=1The zeros of the function are -1 and 1 multiplicity 2 and the y-intercept is 1c. Y 1 X 1 -1 13

d. The function G( x ) ≥ 0 when x ≥ -1 e. The graph is tangent to the x-axis at 1, since 1 is a zero of even multiplicity that is 2. 2. F( x ) = x4 + 2x3 - 5x2 - 6x a. Determine the behavior of the graph b. Get the zeros and the y-intercept of the function. c. Sketch the graph d. Determine the interval where G( x ) ≥ 0Solution: a. The degree of the function is 4 which is an even, it is an even-powered function behaving like a parabola opening upward. Extreme bounds:12 -5 -6 0 -4 -4 8 -12 72 3 -18 721 -2-4 is the largest lower bound.1 2 -5 -6 0 2 28 60 0014 32 is the smallest integral upper bound and a zero.b. Location of zerosx -4 -3 -2 -1 0 1 2F(x) 72 0 -8 0 0 -8 0 14

The table reveals the following information: 1. The zeros of F( x ) are -3, -1, 0 and 2. 2. y-intercept is 0. c. Sketch of the graph. Y X -4 -3 -2 -1 0 1 2 3 Since n = 4, the function F( X ) has three turning points. d. Since n = 4 and an = 1, the extreme left part is decreasing and the extreme right part is increasing. The function F(x) 〉 0 when x 〈 -3, when -1 〈 x 〈 0 or x 〉 2.3. p( x ) = x3 - 4x2 - 4x + 16 a. Determine the behavior of the graph b. Get the zeros and the y-intercept of the function. c. Sketch the graph d. Determine the interval where p( x ) ≥ 0.Solution: a. The degree of the function is 3 which is an odd, it is an odd-powered function behaving like a line that starts from the right. Since the leading coefficient is positive, it is increasing. 15

b. Possible combination zeros: ±1, ±2, ±4, ±8, ±16Extreme bounds: 1 -4 -4 16 5 55 5 1 1 1 215 is the least integral upper bound. 1 -4 -4 16 -2 -2 12 -16 80 1 -6-2 is the greatest integral lower bound and a zero.b. Location of zerosX -2 -1 0 1 2 3 4 5Y 0 15 16 9 0 -5 0 21c. Sketch of the graph. Y 20 15 10 5X -2 -1 0 12 34 5 -5 -10 -15 -20 Since n = 3, the function p (x ) has two turning points. 16

The table reveals the following information: 1. The zeros of p( x ) are -2, 2 and 4. 2. y-intercept is 16. d. The function p( x ) ≥ 0 if 2 ≥ x < 2 or x > 4.4. G( x ) = -x5 + 15x3 + 10x2 - 24x a. Determine the behavior of the graph b. Get the zeros and the y-intercept of the function. c. Sketch the graph Solution: a. The degree of the function is 5 which is an odd, it is an odd-powered function behaving like a line that starts from the left since the leading coefficient is negative -1. The extreme left and right parts are decreasing. b. Possible zeros: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 The theorems on bounds does not apply for G(x) since the leadingcoefficient is negative. x -3 -2 -1 0 1 23 4 y 0 0 20 0 0 80 180 0The table reveals that the following information: 1. The 5 zeros are -3, -2, 0, 1 and 4 2. The y intercept is 0. c. Sketch the graph.17

YX -1 01234 -3 -25. f( x ) = x3 - 3x2 - 5x + 12 using a graphing calculator and approximate its real zeros to the nearest half unit. Steps: To draw the graph: 1. Use the GRAPH Mode input in the function. 2. Press F6 to draw the graph. From your calculator screen you will see the sketch of the graph like this. Y X -3 -2 -1 0 1 2 3 18

To determine the zeros of f( x ) = x3 -3x2 -5x + 12 1. Press F5 ( G-Solv ) 2. Press F1 ( Root ) to display one root. 3. Press the → arrow key ( 3x ) for other roots. • Press F5 ( G – Solv ) to use othe features such as finding the y- intercept and roots. • Observe the behavior of the graphs when it is odd or even. Note from the graph that the real zeros of f( x ) = x3 -3x2 -5x + 12, to thenearest half unit, are -2.0, 1.5 and 3.5.Try this out Graph each of the following polynomials. a. Determine the behavior of the graph b. Get the zeros and the y-intercept of the function. c. Sketch the graph 1. f( x ) = x3 – x2 – 4x +4 2. F( x ) = 2x4 + 9x3 + 11x2 - 4 3. H( x ) = -x4 + 2x3 + 3x2 -2x + 6 4. Graph the function and approximate its real zeros to the nearest tenth using graphing calculator: y = 2x4 + 5x3 - 3x2 - 9x + 1 19

Let’s summarize To graph polynomial function of degree greater than two, evaluate thefunction to determine ordered pairs, then plot the points with this pairs ascoordinates and connect the points to form a smooth curve. The x-coordinates ofthe points where the graph meets the x-axis are the zeros of the function. Theinformation you can get from the given equation of a polynomial function are thedegree and the number of zeros. From the graph, you can determine the numberof x-intercepts, y-intercept and number of turning points.Summary of the characteristics of the graph of polynomial function:an n Properties of the graph Illustration+ Even Comes down from the left, Figure 1 goes up to the right+ Odd Comes up from the left, Figure 2 goes up to the right- Even Comes up from the left, Figure 3 goes down to the right- Odd Comes down from the left, Figure 4 goes down to the rightFigure 1 Figure 2Figure 3 Figure 4 20

What have you learnedA. Indicate whether the graph represents a polynomial function.1. 2. Y YXX 3. 4. Y 5. Y Y X XXB. Complete the table below: Functions Degree Number No. of x- No. of turning of zeros intercepts points1. P(x) =(x-1)(x+2)(x-1)2. y= x(x-2)3. g(x)= x4 – x3- 7x2+x+64. P(x)= 2x5 +5x4-2x3-7x2-4x-12 21

C. For each of the following polynomials, a. determine behavior of the graph. b. get the zeros and the y-intercept of the function. c. sketch the graph d. interval where G(x) ≥ 0 1. P(x) = x4 + x3+ 4x2 + 6x - 12 2. P(x) = -x5 +15x3 + 10x2 - 24x 22

Answer KeyHow much do you knowA.1. linear 2. quadratic 3. quadratic 4. polynomial 5. polynomialB. degree No. of No. of x-intercept No. of turning zeros points12 2 123 2 3 236 3 6 542 6 2 151 2 1 0 1C. a. behaves like a parabola that opens upward Y b. Zeros are -3, −1 , 1 and 3, y-intercept is 0 2 c. f(x) ≥ 0 when x 〈-3, when −1 〈 x 〈 1 or when x 〉 3 X 2Lesson 1 01 2. 3. 11. -8 -3 X -1 0 1 X -1 0 1 Y531 y53 X -1 y -13 YY XX 23

4. (0, -7) 5. (3, 0)X -1 0 1 X12345Y -6 -7 -6 Y41014 Y YX XLesson 2aA. 1. upper bound: -1 2. upper bound: 4 3. upper bound: 6 Lower bound: -4 lower bound: -4 lower bound: -2B. 4. n = 3 and an = 1. The extreme left and right parts of the function are increasing. 5. n = 4 and an = 6. The extreme left part is decreasing and the extreme right is increasing. 6. n = 5 and an = 1. The extreme left and right parts are increasing.C. 7. Imaginary 8. zeros +- zeros zeros 0 039. 24

+ - Imaginary + - Imaginaryzeros zeros zeros zeros zeros zeros 3 2 0 2 2 0Lesson 2B:1. a. The extreme left and right parts of the function are increasing.b. Zeros are 1, -2 and 3 and the y-intercept is 4 c. Yx -3 -2 -1 0 1 2 3y -20 0 6 4 0 0 10 Xd. f(x) ≥0 for all x〉-22. . a. The graph comes from the left and goes up to the right..b. Zeros are -1, -2, 1 , and 2. The y-intercept is -4. 2c. x -2 -1 0 1 2 y 0 0 -4 18 2 Y X 25

3. a. The extreme left part is decreasing and the extreme right part is increasing.b. Zeros are -3, -1, 0, and 2. The y-intercept is 0.c. Yx -4 -3 -2 -1 0 1 2y 72 0 -8 0 0 -8 0 XWhat have you learnedA. 1. not polynomial nos. 2 – 5 are polynomialsB. Degree Number No. of x- No. of Functions of zeros intercepts turning points1. P(x) =(x-1)(x+2)(x-1) 3 3 3 2 2 2 2 12. y= x(x-2) 4 4 4 33. g(x)= x4 – x3- 7x2+x+6 5 5 5 44. P(x)= 2x5 +5x4-2x3-7x2-4x-12C. 1. a. The graph comes from the left and goes up to the right.b. The real zeros are -2 and 1. The y-intercept is -12. Yc. X x -4 -3 -2 -1 0 1 y 220 72 0 -14 -12 0d. P(x) ≥ 0 for all x 〉 1 or x 〈 -2 26

2 a. The extreme left part is decreasing and the extreme right part is increasing.b. Zeros are -3, -2, 0, 1 and 4. The y-intercept is 0c. -3 -2 1 0 1 2 3 4 x y 0 0 20 0 0 80 180 0 Yd. f( x ) ≥ 0 for x 〉 1 X 27

Module 3 Quadratic Functions What this module is about This module is about the zeros of quadratic functions and the roots ofquadratic equations. As you go over this material, you will be able to determinethe zeros of a quadratic function by relating this to the corresponding roots ofquadratic equation. This material will likewise develop your skills in finding theroots of quadratic equations using factoring, completing the square and quadraticformula. Moreover, you will be able to use learned concepts and skills in derivingquadratic function given certain conditions. What you are expected to learn This module is designed for you to: 1. determine the zeros of a quadratic function by relating this to the roots of a quadratic equation. 2. find the roots of quadratic equations by: a. factoring b. completing the square c. quadratic formula 3. derive quadratic functions given: a. zeros of the function b. table of values c. graph How much do you know 1. What are the zeros of f(x) = 4x2 – 64? 2. Solve the equation x2 – 5x + 6 = 0.

3. Use the quadratic formula in solving x2 – x – 3 = 0.4. Determine the roots of 2x2 – 3x + 1 = 0 using the method of completing the square.5. Find a quadratic function whose zeros are -3 and 2.6. Name the quadratic function satisfied by the table below. x -2 -1 0 2 f(x) 12 6 2 07. Determine the zeros of the quadratic function whose graph is given below. Y X ●●8. Determine the quadratic function whose graph is given below. Y X (-1,-3) ● ● (2,-1) ● ( 1,-7)9. What quadratic function has 2 and − 2 as zeros?10. What quadratic function has a vertex at (2, 1) and passes through (3, -1)? 2

What you will do Lesson 1 Determining the Zeros of Quadratic Function from Its Graph The parabolic structure of the graph of a quadratic function allows it tointersect the x-axis in different ways. It may cross the x-axis once, twice or noneat all as shown below. Y X Graphically speaking, the zeros of a quadratic function f(x) are the x-coordinates of the point of intersection of the graph of f(x) and the x-axis, if itdoes exists. In other word, the zeros of a quadratic function are its x-intercepts. Examples: Determine the zeros of the quadratic function whose graph is given below. Y 1. X ●● ● 3

The parabola intersects the x-axis at -2 and 2. Hence, the zeros of thequadratic function represented by the parabola are -2 and 2. 2. Y X● ●● The parabola intersects the x-axis only at 0. Hence, the only zero of thequadratic function represented by the parabola is 0. 3. Y ●● ● X The parabola did not intersect the x-axis. Hence, the quadratic functionhas no real zero. It is possible that the zeros are imaginary.Try this outSet A Determine the zeros of the quadratic function whose graph is given below. 1. Y ● X ●● 4

2. Y X ●● ●3. Y X ●● ●4. Y ●● X●5. Y X ● 5

Set B Y1. ● X● ●2. X ● Y ● ●3. Y X ●4. Y ●X● ● 6

5. Y X ● Set C Y1. X● ● ●2. Y X●3. Y X● 7

4. Y ● X Y5. ● X● ● Lesson 2Determining the Zeros of Quadratic Functions by Factoring The zeros of a quadratic function of the form f(x) = ax2 + bx + c are theroots of the quadratic equation ax2 + bx + c = 0. If the quadratic expression ax2 + bx + c is factorable, factor it and applythe zero property.Examples:Determine the zeros of each quadratic function.1. f(x) = 2x2 + 4xEquate the given function to zero and then solve for x.f(x) = 2x2 + 4x = 0 2x(x + 2) = 0 Factor out factor 2x2x = 0 or x + 2 = 0 Apply the Zero property x = 0 or x = -2 8

Hence, the zeros of f(x) = 2x2 + 4x are 0 and -2.2. g(x) = x2 – 4Equate the given function to zero and then solve for x. g(x) = x2 – 4 = 0 (x – 2)(x + 2) = 0 Factor the difference of two squares x – 2 = 0 or x + 2 = 0 Apply the zero property x = 2 or x = -2Hence, the zeros of g(x) = x2 – 4 are 2 and -2.3. h(x) = 4x2 – 4x + 1Equate the given function to zero and then solve for x. h(x) = 4x2 – 4x + 1 = 0 (2x – 1)2 = 0 Factor the Perfect square trinomial 2x – 1 = 0 Apply the zero property x = 1 2 Hence, the only zero of h(x) = 4x2 – 4x + 1 is 1 . 24. f(x) = x2 + 2x – 8Equate the given function to zero and then solve for x. f(x) = x2 + 2x – 8 = 0 (x – 2)(x + 4) = 0 Factor x – 2 = 0 or x + 4 = 0 Apply the zero property x = 2 or x = -4Hence, the zeros of F(x) = x2 + 2x – 8 are 2 and -4.5. g(x) = -2x2 – x + 3Equate the given function to zero and then solve for x. g(x) = -2x2 – x + 3 = 0 (-2x – 3)(x – 1) = 0 Factor -2x – 3 = 0 or x – 1 = 0 Apply the zero property x = 3 or x = 1 2 Hence, the zeros of G(x) = -2x2 – x + 3 are 3 and 1. 2Try this outFind the zeros of each quadratic function by factoring. 9

Set A 1. f(x) = 5x2 – 5 2. g(x) = 2x2 – 12 3. h(x) = 3x2 – 27 4. f(x) = x2 + 4x + 3 5. g(x) = x2 + 5x + 6 Set B 1. f(x) = 36x2 – 49 2. g(x) = x2 + 3x + 2 3. h(x) = x2 + 7x + 12 4. f(x) = x2 + 5x – 6 5. g(x) = x2 – x – 12 Set C 1. f(x) = x2 – 5x + 6 2. g(x) = x2 + 5x – 14 3. h(x) = x2 – 4x – 21 4. f(x) = 2x2 + x – 1 5. g(x) = 3x2 + 8x + 4 Lesson 3 The Imaginary Numbers It has been mentioned that if a quadratic function has no real zeros, it ispossible that the zeros are imaginary. A non-real number is called imaginary number. The unit imaginary numberis defined as follows: The imaginary number i is a number whose square root is -1. In symbols, i = −1 → i2 =−1Examples: Simplify each of the following: 1. −25 10

( )( )−25= 25(−1)= 25 −1= 5i 2. −12 ( )( )( )−1=2 4(3)(−=1) 4 −1 =3 2i 3 3. −11 ( )( )−11 = (−1)(11) = −1 11 = i 11Try this out Simplify each of the following: Set A 1. −2 2. −4 3. −36 4. −18 5. −20 Set B 1. −81 2. −28 3. −75 4 4. −27 16 5. −31 16 Set C 1. −52 2. −45 3. −19 49 11

4. −41 165. −162 121 Lesson 4Determining the Zeros of Quadratic Functions by Completing the Square Solving for the zeros of quadratic functions is limited to quadraticexpressions that are factorable. However, there is another method that works forquadratic expressions that are factorable or not. This method is called themethod of completing the square. The principle behind the use of completing the square is to produce aperfect square trinomial so that the square root property can be apply.The following steps are suggested:1. Equate the given quadratic function to zero.2. Transpose the constant term.3. If a ≠1, divide both sides of the equation by a.4. Add to both sides of the equation the square of half the coefficient of x.5. Factor the resulting perfect square trinomial.6. Apply the square root property.7. Solve for x.Examples:Find the zeros of each quadratic function by completing the square.1. y = x2 – 4x – 5. Equate y to 0 0 = x2 – 4x – 5 Add + 5 to both sides of the equation x2 – 4x = 5 Add the square of one-half of -4 to x2 – 4x + 4 = 5 + 4 both sides of the equation. (x – 2)2 = 9 Factor x–2 =±3 x=2±3 Take the square root of 9 Solve for x.x = 5 or x = -1Hence, the zeros of y = x2 – 4x – 5 are 5 and -1. 12

2. y = 3x2 + 4x + 1. Equate y to zero 0 = 3x2 + 4x + 1 Add –1 to both sides of the equation 3x2 + 4x = -1x2 + 4 x = −1 Divide both sides by the coefficient 3. 33x2 + 4 x + 4 = −1 + 4 Add the square of one-half of 4 . 3 9 39 3 x + 2 2 = 1 Factored 3 9 x+ 2 = ±1 Take the square root of 1 33 9 x = −2±1 Solve for x. 33x = − 1 or x = -1 3Hence, the zeros of y = 3x2 + 4x + 1 are − 1 and -1. 33. y = x2 – 2x – 5 Equate y to 0. 0 = x2 – 2x – 5 Add + 5 to both sides of the equation. x2 – 2x = 5 Add the square of one-half of -2 to x2 – 2x + 1 = 5 + 1 both sides of the equation. (x – 1)2 = 6 Factor x–1 = ± 6 x = 1± 6 Take the square root of 6. Solve for x.x= 1 + 6 or x= 1 − 6Hence, the zeros of y = x2 – 2x – 5 are 1 + 6 and1 − 6 .4. y = 2x2 – x + 40 = 2x2 – x + 4 Equate y to zero. Add – 4 to both sides of the equation.2x2 – x = -4 Divide both sides by the coefficient 2.x2 − 1 x = −2 Add the square of one-half of − 1 to 2 2x2 − 1 x + 1 = −2 + 1 both sides of the equation.2 16 16 Factor x − 1 2 = − 31 4 16 Take the square root of − 31 . 16x−1 = ± i 31 4 4 13

X = 1 ± i 31 (The value of x is solved.) 44 x = 1 + i 31 or x = 1 − i 31 44Hence, the zeros of y = 2x2 – x + 4 are 1 + i 31 and 1 − i 31 . 44Try this out:Find the zeros of each quadratic function by completing the square.Set A1. f(x) = x2 – x – 22. g(x) = x2 + 6x + 43. h(x) = x2 – 7x + 124. f(x) = 2x2 – 3x + 15. g(x) = 3x2 + 5x – 2Set B1. f(x) = x2 – 10x + 212. g(x) = x2 + 4x – 13. h(x) = x2 – 8x + 34. y = x2 + 6x – 55. y = 2x2 + 2x + 3Set C1. y = -x2 – 3x + 42. y = x2 – 2x – 53. y = x2 + x – 14. y = 2x2 + 2x – 75. y = 3x2 + 6x + 1 Lesson 5 Determining the Zeros of Quadratic Functions by Using the Quadratic Formula Suppose the general quadratic equation ax2 + bx + c = 0 is solved by themethod of completing the square. 14

ax2 + bx + c = 0 Add –c to both sides of the equation ax2 + bx = -c Divide both sides by the coefficient 2. x2 + b x = −c Add the square of one-half of b to aa a x2 + b x + b2 = −c + b2 both sides of the equation. a 4a2 a 4a2 Factor  x + b 2 = b2 − 4ac Take the square root of the left side. 2a 4a 2 Solve for x. x + b = ± b2 − 4ac 2a 2a Combine the fractions x = − b ± b2 − 4ac 2a 2a x = −b ± b2 − 4ac 2aThe last equation is referred to as the quadratic formula.Examples:Find the zeros of each quadratic function by using the quadratic formula.1. y = x2 + 2x – 3. First, equate the given quadratic function to 0. Identify the values of a, b,and c. Then, substitute these values to the quadratic formula and solve for x. a = 1 b = 2 c = -3 x = −b ± b2 − 4ac 2a −2 ± 22 − 4(1)(−3) x= 2(1) x = −2 ± 4 + 12 2 x = −2 ± 16 2 x = −2 ± 4 2 x = 1 or x = -3Hence, the zeros of y = x2 + 2x – 3 are 1 and -3. 15

2. y = 2x2 – x – 1. First, equate the given quadratic function to 0. Identify the values of a, b,and c. Then, substitute these values to the quadratic formula and solve for x. a = 2 b = -1 c = -1 x = −b ± b2 − 4ac 2a −(−1) ± (−1)2 − 4(2)(−1) x= 2(2) x = 1± 1+8 4 x = 1± 9 4 x = 1±3 4 x = 1 or x = − 1 2 Hence, the zeros of y = 2x2 – x – 1 are 1 and − 1 . 2 3. y = x2 – 5x + 2. First, equate the given quadratic function to 0. Identify the values of a, b,and c. Then, substitute these values to the quadratic formula and solve for x. a = 1 b = -5 c = 2 x = −b ± b2 − 4ac 2a −(−5) ± (−5)2 − 4(1)(2) x= 2(2) x = 5 ± 25 − 8 4 x = 5 ± 17 4 x = 5 + 17 1 or x = 5 − 17 44 Hence, the zeros of y = x2 – 5x + 2 are 5 + 17 and 5 − 17 . 44 16

4. y = 3x2 – 3x + 2. First, equate the given quadratic function to 0. Identify the values of a, b,and c. Then, substitute these values to the quadratic formula and solve for x. a = 3 b = -3 c = 2 x = −b ± b2 − 4ac 2a −(−3) ± (−3)2 − 4(3)(2) x= 2(3) x = 3 ± 9 − 24 6 x= 3±i 15 6 x = 3 + i 15 1 or x = 3 − i 15 66Hence, the zeros of y = 3x2 – 3x + 2 are 3 + i 15 and 3 − i 15 . 66Try this outFind the zeros of each quadratic function by using the quadratic formula.Set A1. f(x) = x2 + 6x + 52. g(x) = 3x2 + 6x – 23. h(x) = 4x2 – 3x – 24. F(x) = 2x2 – x – 35. G(x) = 4x2 – 4x – 3Set B1. f(x) = x2 + x – 202. g(x) = x2 + 4x – 123. h(x) = 3x2 + 2x - 14. y = 2x2 + x – 55. y = 3x2 – x – 2Set C1. y = -4x2 + 2x – 32. y = -4x2 + 8x + 3 17

3. y = x2 + x + 3 4. y = 2x2 + 4x – 7 5. y = 3x2 + 6x – 2 Lesson 6 Deriving a Quadratic Function, given the Zeros of the Function Recall that the zeros of a quadratic function are also the roots of thecorresponding quadratic equation. If the zeros of a quadratic function y are x1and x2, then x – x1 and x – x2 are factors of the quadratic expression. Thus, thecorresponding quadratic equation is (x – x1)(x – x2) = 0and the quadratic function is y = (x – x1)(x – x2)Examples: Derive the quadratic function, given the zeros of the function: 1. -2, 3 Let y be the quadratic function. If -2 and 3 are the zeros of y, then x – (-2)and x – 3 are factors of y. Hence, y = [x – (-2)](x – 3) 2. − 1 ,1 y = (x + 2)(x – 3) 2 y = x2 – x – 6 Let f(x) be the quadratic function. If − 1 and 1 are the zeros of f(x), then 2x −  − 1  and x – 1 are factors of f(x). The corresponding quadratic equation is 2 x −  − 1  (x – 1) = 0 2  x + 1  (x – 1) = 0 2 (2x + 1)(x – 1) = 0 18

2x2 – x – 1 = 0. Hence, the quadratic function is f(x) = 2x2 – x – 1. 3. ± 2 Let g(x) be the quadratic function. If ± 2 are the zeros of g(x), then x –( )2 and x – − 2 are factors of g(x). The corresponding quadratic equation is( ) ( )x − 2 x − − 2  = 0(x − 2)(x + 2) = 0 ( )2 x2 – 2 = 0 x2 – 2 = 0 Hence, the quadratic function is g(x) = x2 – 2. 4. 1 ± i Let h(x) be the quadratic function. If 1 ± i are the zeros of g(x), then x –(1 + i) and x – (1 − i) are factors of h(x). The corresponding quadratic equation is [x – (1 + i) ][x – (1 − i) ] = 0 (x – 1 + i)(x – 1 – i) = 0 [(x – 1) + i][(x – 1) + i] = 0 (x – 1)2 – (i)2 = 0 x2 – 2x + 1 – (-1) = 0 x2 – 2x + 1 + 1 = 0 x2 – 2x + 2 = 0 Hence, the quadratic function is h(x) = x2 – 2x + 2.Try this out Derive the quadratic function, given the zeros of the function: Set A 1. 1, 3 2. -3, 4 3. 4, 6 19