Syllabus Objective: 2.9 The student will sketch the graph of a polynomial, radical, or rational function.

Transcription

1 Precalculus Notes: Unit Polynomial Functions Syllabus Objective:.9 The student will sketch the graph o a polynomial, radical, or rational unction. Polynomial Function: a unction that can be written in the orm n n1 an an 1... a a1 a0, where n is a nonnegative integer and an 0 Note: This is called Standard Form when the eponents o descend The graph o a polynomial unction is a continuous curve with smooth round turns. Degree o a Polynomial Function: the largest eponent, n, o Leading Coeicient o a Polynomial Function: the coeicient o the irst term when written in standard orm, a n Constant Term: the numerical term, a 0 Classiying Polynomial Functions I. Number o Terms 1 term = monomial terms = binomial terms = trinomial 4 or more terms = polynomial Degree II. Degree Name (Degree) Standard Form Eample Classiication o Eample y Constant Monomial y Linear 0 Constant a 1 Linear a b Quadratic a b c Cubic a b c d 4 Quartic 4 a b c d e 5 Quintic 5 4 a b c d e Monomial y 5 7 Quadratic Trinomial y 5 Cubic Binomial 4 y 9 1 Quartic Polynomial 5 y Quintic Monomial Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

2 Precalculus Notes: Unit Polynomial Functions E1: Which o the ollowing are polynomial unctions? State the degree and the leading coeicient or eplain why it is not a unction. g a) 8 17 NOT a polynomial unctions because n 8. h Degree = 5, Leading Coeicient = 7 (quintic trinomial) b) 5 Linear Functions A line in the Cartesian plane is a linear unction i and only i it is a slant line (not vertical because it ails the vertical line test; not horizontal because it is a constant unction). The slope, or rate o change o a linear unction is constant. b a rise y1 y y y1 change in y y dy Average Rate o Change = = m = b a run change in d 1 1 E: Write an equation o the linear unction that has and 4 7 m 5 Find the slope o the line Write equation using point-slope orm. y or y4 Modeling with Linear Functions Correlation: Positive Negative Relatively No Linear Correlation Slope: Positive Negative no line E: Use the data to write a linear model or the Women s Olympic 100 meter times below. Describe the correlation and predict the time in 010. Year (19 ) Time (sec) : y sec y Page o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

3 Precalculus Notes: Unit Polynomial Functions Quadratic Function: a polynomial unction o degree Graph is u-shaped, called a parabola Parabolas are symmetric with respect to a line called the ais o symmetry Parabolas have a verte, which is the point on the parabola where it intersects the ais o symmetry, and is the maimum or minimum o the quadratic unction General Form: b a b c Ais o Symmetry: a b b 4ac o -intercepts can be ound using the quadratic ormula: a hk Ais o Symmetry: h o In both orms, i a 0, the parabola opens UP; i a 0, the parabola opens DOWN Verte Form: ah k Verte:, Graphing a Parabola E4: Write the equation in verte orm. Find the verte and ais o symmetry. Then graph the unction. 4 1 Complete the square to write in verte orm: y y y 1 y Alternate Method: Verte: 1, Ais o Symmetry: 1 y Writing the Equation o a Parabola Substitute the verte or hk, in verte orm. Substitute the given point or, y and solve or a. Page o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

4 Precalculus Notes: Unit Polynomial Functions E5: Write an equation or the parabola with verte hk,, 4 a 4 y, 4, 6 4 6a 4 4 6a4 a, 4 that passes through the point 4, 6. Average Rate o Change (recall): b b a a E6: Find the average rate o change o rom to Ave Rate o Change = 4 Vertical Free-Fall Motion: ormula or the position o an object in ree all 1 s t gt v0t s0 g t/sec 9.8 m/sec (on Earth) s = position g = acceleration due to gravity v 0 = initial velocity s 0 = initial position 6 E7: A lare is shot straight up rom a ship s bridge 75 eet above the water with an initial velocity o 76 t/sec. How long does it take to reach its maimum height? How long does it take to reach 16 eet? s0 75 v0 76 g t/sec s t t t s t t t The lare reaches it ma height at the verte o the parabola. It takes.75sec or the lare to reach its maimum height. Note: This is NOT the graph o the path o the lare. The graph relates the position o the lare to time. The lare reaches 16 eet when 16 16t 76t 75. It reaches 16 eet at sec &.75 sec Page 4 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

5 Precalculus Notes: Unit Polynomial Functions t t t t t t Algebraic Method: t t t t t t 11 t 4t11 0 t, 4 You Try: Write the unction in verte orm. Find the verte and ais o symmetry. Then graph. QOD: Using a table o values, how can you determine whether they have a linear or quadratic relationship? Page 5 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

6 Precalculus Notes: Unit Polynomial Functions Syllabus Objectives:.1 The student will graph relations or unctions, including real-world applications..9 The student will sketch the graph o a polynomial, radical, or rational unction. a Power Function: a unction that can be written in the orm k, where k and a are non-zero constants k = constant o variation/proportion a = power, a 0 Direct Variation: a 0 Inverse Variation: a 0 Note: Variation is assumed direct unless it speciies inversely. Direct & Inverse Variation E1: Write the statement as a power unction: The area o a circle varies as the square o the radius. Implied direct variation. A = Area, r = radius A k r We know that k, so A r E: y varies inversely as q. y when q 6. Find q when y 15. k Inverse variation: y kq 1 or y q k Solve or k using the given inormation: k 1 6 Use k and y 15 to ind q. Identiying Power Functions q1 q q 15 5 E: Determine i each unction is a power unction. I yes, state k and a. (c is a constant.) 0.5 Yes, k 0.5, a a) b) g No, because a 0 c) h No, because the base is not a variable d) g e) ) h c Yes, k, a 1 c Yes, k c, a Yes, k c, a 1/ Monomial Function: a k or k, where k is a constant and n is a positive integer Graphing Power Functions: Teachers Allow students time to eplore the graphs o various power unctions on the graphing calculator, including positive, negative, and rational values o a. Page 6 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

7 Precalculus Notes: Unit Polynomial Functions E4: Graph and analyze the unction: Domain:,0 0, Range:,0 0, Continuity: Continuous Function. Ininite discontinuity at 0 Increasing/Decreasing: always decreasing Symmetry: odd; Boundedness: unbounded Etrema: none Asymptotes: y 0, 0 lim 0, lim 0 End Behavior: y Und y Modeling with Power Functions Newton s Law o Cooling E5: The rate at which an object cools varies as the dierence between its temperature and the temperature o the surrounding air. When a 70 C steel plate is placed in air that is at 0 C, it is cooling at 50 C per minute. How ast is it cooling when its temperature is 100 C? R = rate o cooling, T 0 = initial temp, T S = surrounding temp: R kt0 TS 1 Solve or k: T0 70, TS 0, R k k k 5 Answer the question: R C / min 5 You Try: The electrical resistance o a wire varies directly as its length and inversely as the square o its diameter. 100 m o a wire with diameter 6 mm has resistance 1 ohms. 80 m o a second wire o the same material has resistance 15 ohms. Find the diameter o the second wire. QOD: Are all power unctions monomial unctions? Are all monomial unctions power unctions? Eplain. Page 7 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

8 Precalculus Notes: Unit Polynomial Functions Syllabus Objective:.9 The student will sketch the graph o a polynomial, radical, or rational unction. a a... a a a, a 0 n n1 Polynomial Function (recall): Review (i necessary): Degree Name (Degree) n n1 1 0 Standard Form Eample Classiication o Eample y Constant Monomial y Linear 0 Constant a 1 Linear a b Quadratic a b c Cubic a b c d 4 Quartic 4 a b c d e 5 Quintic 5 4 a b c d e y-intercept: occurs at 0. Functions have only 1 y-intercept. n Monomial y 5 7 Quadratic Trinomial y 5 Cubic Binomial 4 y 9 1 Quartic Polynomial 5 y Quintic Monomial 4 7 E1: Describe how to transorm the graph. Name the y-intercept. Parent Function = cubic unction y Vertical stretch by a actor o Relect over the -ais Shit let 4 units Shit up 7 units y-intercept: y E: Graph the unction and list its characteristics. Domain: All Reals Range: All Reals Increasing:,0 1, Decreasing: 0,1 Symmetry: None Boundedness: None Etrema: Local Ma: y 0 at 0, Local Min: y 1 at 4 End Behavior: lim y ; lim 0 0, Zeros: Page 8 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

9 Precalculus Notes: Unit Polynomial Functions Zeros (-intercepts, Roots) o Polynomial Functions: nth-degree polynomials have at most n 1 etrema and n zeros Graph: 5 4. E: Find the zeros and etrema o the unction 4 Limit: the value that Notation: lim lim a a lim a Zeros:, 1,1, Etrema:.5 (occurs twice), 4 approaches as approaches a given value The limit as approaches a o The limit o The limit o. (From both sides.) as approaches a rom the let. as approaches a rom the right. End Behavior o Polynomial Functions Odd Degree: I the leading coeicient is positive, then lim & lim. I the leading coeicient is negative, then lim & lim. Even Degree: I the leading coeicient is positive, then lim & lim. I the leading coeicient is negative, then lim & lim. E4: Indicate i the degree o the polynomial unction shown in the graph is odd or even and indicate the sign o the leading coeicient. a) Odd degree; Negative leading coeicient b) Even degree; Positive leading coeicient Page 9 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

10 Precalculus Notes: Unit Polynomial Functions Zeros o Polynomial Functions Multiplicity: repeated zeros; I a polynomial unction has a actor o then c is a zero o multiplicity m o. Odd Multiplicity: crosses the -ais at c; changes signs c m m 1, and not c Even Multiplicity: kisses or is tangent to the -ais at c; doesn t change signs E5: Sketch the graph o g 1 : multiplicity 1. Describe the multiplicity o the zeros. : multiplicity y-intercept: End behavior (odd degree o 5, positive leading coeicient): lim y ; lim, y Watch or hidden behavior: E6: Graph the unction in a Standard window and ind the zeros is degree, so the end behavior is lim, so we know it must cross the -ais again to the right. New window: Zeros are 1,, 0 Intermediate Value Theorem (Location Principle): I a and b are real numbers with a b, and i is continuous on ab,, then takes on every value between a and b. Thereore, i a and b have opposite signs, then c 0 or some number c in, ab. Page 10 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

11 Precalculus Notes: Unit Polynomial Functions 1, 0 : E7: Use the Intermediate Value Theorem to show in these intervals: 1, 0 and 1, (negative) (positive) changes sign (negative to positive) in the interval 1, (positive) 1, : (negative) changes sign (positive to negative) in the interval has zeros, so must have at least one zero in 1, 0 1,, so must have at least one zero in. 1,. Application o Polynomial Functions E8: You cut equal squares rom the corners o a by 0 inch sheet o cardboard to make a bo with no top. What size squares would need to be cut or the volume to be 00 cubic inches? Deine as the length o the sides o the squares. Write a ormula or the volume o the bo. V 0 Solve the equation 00 0 by graphing. Squares with lengths o approimately 0.49 or inches should be cut. You Try: Sketch the graph o the polynomial unction. 4 5 QOD: How many zeros can a unction o degree n have? Eplain your answer. 1 Page 11 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

12 Precalculus Notes: Unit Polynomial Functions Syllabus Objectives:. The student will calculate the intercepts o the graph o a given relation. Review: Long Division 4, Remainder is 19, so the quotient is Long Division o Polynomials (same process!) E1: Find the quotient. Note: Every term o the polynomial in the dividend must be represented. Since this polynomial is missing an term, we must include the term Solution: Remainder Theorem: I is divided by k, then the remainder, r k. E: Find the remainder without using division or r divided by Note: Compare your answer to the long division problem above. Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

13 Precalculus Notes: Unit Polynomial Functions Factor Theorem: k is a actor o i and only i k 0. E: Determine i is a actor o 5 1 without dividing. Show that r , so by the Factor Theorem, is a actor o 5 1 :. Synthetic Substitution: E4: Find 4 i using synthetic substitution. Using the polynomial in standard orm, write the coeicients in a row. Put the -value to the upper let Bring down the irst coeicient, then multiply by the -value multiply 6 Add straight down the columns, and repeat The number in the bottom right is the value o 5 E 5: Find i using synthetic substitution. This polynomial unction is in standard orm, however it is missing two terms. We can rewrite the unction as to ill in the missing terms. This also means that, 17 is an ordered pair that would be a point on the graph. And the remainder when dividing 17 So : () 47 by is 17. a a... a a a n 1 Recall: n n1 1 0 Possible Rational Zeros o a Polynomial Function = actors o constant actors o leading coeicient a 0 a n Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

14 Precalculus Notes: Unit Polynomial Functions E6: Find the possible rational zeros o ( ) 7 1 Step 1: The leading coeicient is 1. 1 is the only actor o 1. Step : The constant is 1. All o the actors o 1 are 1,,, 4, 6, 1. Step : List the possible actors ,,,,,and *I we tested or actual zeros using synthetic substitution we would ind that and 4 are zeros. This also means that 5 could not be a zero, 7 could not be a zero, 1 could not be a zero,... Descartes Rule o Signs: The number o real zeros equals the number o sign changes in or that number less a multiple o. The number o negative real zeros equals the number o sign changes in or that number less a multiple o. E7: Use Descartes Rule o Signs to determine the possible number o positive and negative real zeros o the unction 4 5. Number o sign changes o : Number o sign changes o : sign change 4 5 sign changes There is 1 possible positive real zero and or 0 possible negative real zeros. Upper and Lower Bound Rules: used to determine i there is no zero larger or smaller than a number c, i the leading coeicient is POSITIVE an 0, using synthetic division or c Upper Bound: c is an upper bound or the real zeros o i c 0 and every number in the last c using synthetic division has signs that are all nonnegative. line o Lower Bound: c is a lower bound or the real zeros o i c 0 and every number in the last line c using synthetic division has signs that are alternately nonnegative and o nonpositive. Page 14 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

15 Precalculus Notes: Unit Polynomial Functions : E8: Veriy that all o the real zeros o the unction interval, lie on the c 0 and signs are all nonnegative, so c is an upper bound o the zeros : c 0 and signs alternate, so c is an lower bound o the zeros Conclusion: All real zeros lie on the interval,. Finding the Real Zeros o a Polynomial: 4 5 E9: Find all real solutions o the polynomial equation. Note: There is one sign change, so there is one positive real zero. 1 1 Possible Rational Zeros:,,, 1 1 Synthetic Substitution (try 1): is a zero, so we can write Factor the quadratic: Solutions: 1,, Note: Remember, i 1 is a zero, 1 is a actor 1 10 ; You Try: Find the real zeros o 4 QOD: How is the Factor Theorem connected to the Remainder Theorem? Page 15 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

16 Precalculus Notes: Unit Polynomial Functions Syllabus Objectives: 1.7 The student will solve problems using the Fundamental Theorem o Algebra. 1.8 The student will calculate the comple roots o a given unction. Fundamental Theorem o Algebra A polynomial unction o degree n has n comple zeros (some zeros may repeat) Comple zeros come in conjugate pairs Review: a bi conjugate a bi A polynomial unction with odd degree and real coeicients has at least one real zero Teacher Note: Have students discuss why. E1: Find all the zeros o the unction and write the polynomial as a product o linear actors Step One: Check or any rational zeros. actors o Possible rational zeros:,, actors o Use synthetic substitution: So is a zero, and is a actor. Step Two: Rewrite the last line as a polynomial and ind the zeros o the polynomial. 0 Factor by grouping: i Step Three: List the zeros. Check that you have the correct number according to the FTA., ii, The polynomial was degree 4, and there are 4 zeros ( is repeated). Note that i and i are comple conjugates. Step Four: Write the polynomial as a product o linear actors. i i E: Find a polynomial unction with real coeicients in standard orm that has the zeros and 1 i. Write the polynomial as a product o linear actors using the zeros. 1i1 i Note: I 1 i is a zero, then its comple conjugate, 1 i, must also be a zero. Page 16 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

17 Precalculus Notes: Unit Polynomial Functions Write the polynomial in standard orm. 1 1 i i 5 10 Irreducible Quadratic Factor: A quadratic actor o a polynomial unction is irreducible over the reals i it has real coeicients but no real zeros. E: Write the polynomial as the product o actors that are irreducible over the reals. Then 5 4 write the polynomial in completely actored orm Step One: Check or any rational zeros Possible rational zeros:,,,,,,,,,,, Use synthetic substitution: Step Two: Rewrite the last line as a polynomial and ind the zeros o the polynomial i 4 0 Product o actors irreducible over the reals: Completely actored orm: 1 5 5i i E4: Use the given zero to ind ALL the zeros o the unction. Zero = 1 i ; Note: By the FTA, we know that must have 4 zeros. Since 1 i is a zero, we know that 1 i (its conjugate) must also be a zero. Two actors o are: 1 i and 1 i 1i 1i Multiply: Page 17 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

18 Precalculus Notes: Unit Polynomial Functions Use long division: Zeros o 4 :,,1 i,1 i Zeros o 0 You Try: Find a polynomial unction with real coeicients in standard orm that has the zeros, and 1 with multiplicity. QOD: Is the polynomial unction ound in the You Try unique? Eplain your answer. Page 18 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

19 Precalculus Notes: Unit Polynomial Functions Syllabus Objectives: 1.9 The student will solve rational equations in one variable..9 The student will sketch the graph o a polynomial, radical, or rational unction. Rational Function: the ratio o two polynomial unctions r, g 0 g Domain o a Rational Function: the set o all real numbers ecept the zeros i its denominator Recall: Reciprocal Function y 1 Rational Functions that are Transormations o the Reciprocal Function 1. g. g E1: Describe the transormations on 1 1 Domain: All reals, 1 1 Relect over -ais. g Translate let by 1 4. g. State the domain. g 1 1 Using Long Division (used when variable is in the numerator and denominator) Long division: E: Describe the transormations on State the domain. g g g Translate down. g. g Vertical stretch by g 1 Vertical stretch by Translate down Relect over -ais 11 5 Translate right 5 Page 19 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

20 Precalculus Notes: Unit Polynomial Functions Recall: Limit: the value that Notation: lim lim a a lim a approaches as approaches a given value The limit as approaches a o The limit o The limit o. (From both sides.) as approaches a rom the let. as approaches a rom the right. E: Evaluate the limits based on the graph. 1) lim 1 ) lim 1 ) lim 4) lim lim 1 lim 1 lim lim n a n... End Behavior Asymptotes o m b m... I n m: horizontal asymptote y 0 an I n m: horizontal asymptote y (ratio o the leading coeicients) b n I n m: No horizontal asymptote. End behavior is the unction that is the result ater perorming the division on. Note: I the n is one more than m, then the unction describing end behavior is linear. This is called a slant asymptote. Page 0 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

21 Precalculus Notes: Unit Polynomial Functions E4: Describe the end behavior o the ollowing unctions. 1) ) y 1 1 Long Division: degree o the numerator > degree o the denominator No horizontal asymptote, no slant asymptote y End Behavior: Approaches the graph o the unction 6 y 1 Lead coeicient o numerator = 6 y 4 5 degree o the numerator = degree o the denominator Lead coeicient o denominator = 1: End Behavior: horizontal asymptote o 6 1 y y 1 Graphing Rational Functions: To sketch the graph o a rational unction, ind the domain, asymptotes, and intercepts. The plot points in each region created by the asymptotes to locate the curve and use the asymptotes to guide the sketching o the curve. E5: Graph g 1. Domain: All reals, 1 Vertical Asymptote: ,1 -intercepts (zeros o numerator): y-intercepts ( g 0 ): g 0 0 End Behavior: No horizontal asymptote; long division results in Slant Asymptote: y Note: Plot a ew points to graph the lower part o the curve. y g lim Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

22 Precalculus Notes: Unit Polynomial Functions 86 E6: Graph g. 4 Domain: All reals,, Vertical Asymptotes:, , 1 -intercepts (zeros o numerator): y-intercepts ( g 0 ): 0 y g End Behavior: horizontal asymptote y Removable Discontinuity: a rational unction has a hole in the graph i one or more o the linear actors in the denominator can be cancelled with one o the actors in the numerator. E7: Sketch the graph o Domain: All reals,, 0 Vertical Asymptotes: g g 18 9 Note: is NOT a vertical asymptote. It is a hole. -intercepts (zeros o numerator): 0 0 0, y-intercepts ( g ): None End Behavior: horizontal asymptote y y You Try: Sketch the graph o y. Describe all relevant eatures. 5 QOD: Can a unction cross a vertical asymptote? Eplain. Can a unction cross a horizontal asymptote? Eplain. Page o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

23 Precalculus Notes: Unit Polynomial Functions Syllabus Objective: 1.9 The student will solve rational equations in one variable. Solving a Rational Equation: 1. Multiply every term by the LCD to wipe out the ractions.. Solve the resulting equation.. Check or etraneous solutions in the original equation, which can result when multiplying by a variable epression E1: Solve the equation. y y y 4 Factor to ind LCD: Multiply by LCD: y y y y y y y 1 y y 1 y LCD = y y y y yy Solve the resulting equation: y y4 y 4 y Check or etraneous solutions: y makes one or more o the denominators in the original equation equal zero, so it is etraneous. Thereore, there is NO SOLUTION. E: Solve the equation: 54 8 This equation is a proportion, so we will cross multiply Solve the equation by actoring: , Check or etraneous solutions: neither solution results in a denominator o zero in the original equation Page o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

24 Precalculus Notes: Unit Polynomial Functions 4 E: Find the solution set o the equation Factor to ind LCD: Multiply by LCD: 5 5 Solve the resulting equation: 5 LCD = ,5 5 Check or etraneous solutions: does not make one or more o the denominators in the original equation equal zero, so it is a solution. 5 does make one or more o the denominators in the original equation equal zero, so it is an etraneous solution. Solution: 4 You Try: Solve the equation 6. QOD: Eplain two ways you can check your solution graphically when solving a rational equation. Sample SAT Question(s): Taken rom College Board online practice problems. 1. The projected sales volume o a video game cartridge is given by the unction sp 000 p a, where s is the number o cartridges sold, in thousands; p is the price per cartridge, in dollars; and a is a constant. I according to the projections, 100,000 cartridges are sold at $10 per cartridge, how many cartridges will be sold at $0 per cartridge? (A) 0,000 (B) 50,000 (C) 60,000 (D) 150,000 (E) 00,000. a c 1 S. I 0 a bcd e in the equation given, then the greatest increase in S b d e would result rom adding 1 to the value o which variable? (A) a (B) b (C) c (D) d (E) e Page 4 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

25 Precalculus Notes: Unit Polynomial Functions. I is deined or all positive numbers a and b by a b = ab, then 10 = a b (A) 5 (B) 5 (C) 5 (D) 0 (E) 0 Page 5 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

26 Precalculus Notes: Unit Polynomial Functions Syllabus Objective:.10 The student will solve and graph inequalities in one variable. Solving an Inequality: 1. Find the zeros o the corresponding equation and the zeros o the denominators o any rational epressions (called critical values).. Make a sign chart (choose a test value) to ind the correct interval solutions. E1: Solve the inequality 4 4 Critical Values: Sign Chart: ,,8 + + Solution: 5,,8 Note: I the inequality was , the solution would be 5,8. E: Solve the inequality: 0 Critical Values: 0 0 not in domain + Sign Chart: Note: is not in the domain o the original unction. Solution:, 1 1 E: Solve the inequality Rewrite the inequality: Critical Values: Sign Chart: 1 Solution:,, 1 Page 6 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter

27 Precalculus Notes: Unit Polynomial Functions Solving an Inequality Graphically E4: Solve graphically: 41 0 Find the zeros (watch or hidden zeros!): Note: The nd and rd graphs were created by zooming in to the local minimum times! Solution:, ,0.5 Application Problem E5: A cylindrical can must have a volume o 58 cubic inches. What dimensions can you use i you must keep the surace area below 100 square inches? Volume: V r h 58 Surace Area: SA r rh 100 Solve or h in the volume equation and substitute into the surace area inequality h r r 100 r 100 r r r Solve by graphing: Solution: Radius must be between 1.97 and.179 inches. Height must be between 1.87 and inches. 1 You Try: Solve the inequality QOD: Eplain why you need only choose one test point within the intervals created by the critical value. Page 7 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter