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
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 1 7 11 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. 7 7 5 5 4. 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 ) 48 56 64 7 80 88 96 Time (sec) 11.9 11.5 11.4 11.1 11.1 10.5 10.9 010: y 100.0101.9 10.5 sec y 0.0 1.9 Page o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter
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: 4 1 1 y y y 1 y 1 1 1 1 Alternate Method: 41 1 1 1 1 1 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
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 4. 4 16 4 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 1 76 75 16 76 75 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
Precalculus Notes: Unit Polynomial Functions 16 16 76 75 16 76 88 0 4 19 0 t t t t t t 4 11 8 0 4 11 4 11 0 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
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 1 1 44 15 15q1 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
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 1 8 1 1 0 Und 1 1 1 8 1 8 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 50 50 k70 0 50 50k k 5 Answer the question: 1 100 0 16 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
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 04 7 11 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
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
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 8 1 16 Watch or hidden behavior: E6: Graph the unction in a Standard window and ind the zeros. 1 8 60 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
Precalculus Notes: Unit Polynomial Functions 1, 0 : E7: Use the Intermediate Value Theorem to show in these intervals: 1, 0 and 1,. 1 1 1 55 1 18 1 40 45 (negative) 0 1 0 55 0 18 0 40 40 (positive) changes sign (negative to positive) in the interval 1, 0 1 1 1 55 1 18 1 40 15 (positive) 1, : 1 55 18 40 48 (negative) changes sign (positive to negative) in the interval 1 55 18 40 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 9.558 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
Precalculus Notes: Unit Polynomial Functions Syllabus Objectives:. The student will calculate the intercepts o the graph o a given relation. Review: Long Division 4,581 1894 4581 05 184 18 07 111 9 19 Remainder is 19, so the quotient is 19 1894 Long Division o Polynomials (same process!) 4 8 116 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 0. 11 88 4 8 0 116 4 11 0 11 11 99 88 6 58 Solution: 11 88 88 64 58 Remainder Theorem: I is divided by k, then the remainder, r k. E: Find the remainder without using division or 4 4 4 8 11 6 r 8 11 6 divided by. 8 11 6 58 Note: Compare your answer to the long division problem above. Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter
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 0 51 0, so by the Factor Theorem, is a actor o 5 1 :. Synthetic Substitution: E4: Find 4 i 5 6 1 using synthetic substitution. Using the polynomial in standard orm, write the coeicients in a row. Put the -value to the upper let. 1 5 6 1 Bring down the irst coeicient, then multiply by the -value. 1 5 6 1 multiply 6 Add straight down the columns, and repeat. 1 5 6 1 6 10 0 48 5 15 4 47 The number in the bottom right is the value o 5 E 5: Find i. 7 11 using synthetic substitution. This polynomial unction is in standard orm, however it is missing two terms. We can rewrite the unction as 5 4 0 7 0 11 to ill in the missing terms. This also means that, 17 is 1 0 7 0 11 an ordered pair that would be a 9 1 4 16 point on the graph. And the 1 7 14 4 17 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
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 - 1 4 6 1,,,,,and 1 1 1 1 1 1 *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 : 4 5 1 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
Precalculus Notes: Unit Polynomial Functions : E8: Veriy that all o the real zeros o the unction interval,. 4 5 6 0 0 10 15 8 4 5 lie on the c 0 and signs are all nonnegative, so c is an upper bound o the zeros : 4 5 9 15 0 5 10 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): 1 1 4 5 7 7 0 1 is a zero, so we can write Factor the quadratic: Solutions: 1,,1 1 7 0. 1 1 0 Note: Remember, i 1 is a zero, 1 is a actor 1 10 ; 0 7 8 14 8. 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
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. 6 10 6 9 4 Step One: Check or any rational zeros. actors o 9 1 9 Possible rational zeros:,, actors o 1 1 1 1 1 6 10 6 9 Use synthetic substitution: 9 9 1 1 0 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: 1 0 1 0 10 0 1 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
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. 60 0 Step One: Check or any rational zeros. 1 4 5 10 0 1 4 5 10 0 Possible rational zeros:,,,,,,,,,,, 1 1 1 1 1 1 Use synthetic substitution: 1 1 1 60 0 1 0 1 0 0 0 0 60 0 Step Two: Rewrite the last line as a polynomial and ind the zeros o the polynomial. 1 4 1 60 4 0 1 4 0 4 0 0 5 4 0 5 0 5 i 4 0 Product o actors irreducible over the reals: 1 5 5 4 Completely actored orm: 1 5 5i i E4: Use the given zero to ind ALL the zeros o the unction. Zero = 1 i ; 4 6 6 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
Precalculus Notes: Unit Polynomial Functions Use long division: Zeros o 4 :,,1 i,1 i 6 6 4 6 6 6 6 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
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 5 4 15 11 11 5 11 5 1. State the domain. g g 11 5 1. g Translate down. g. g Vertical stretch by 11 4. g 1 Vertical stretch by Translate down 4 5 11 5 Relect over -ais 11 5 Translate right 5 Page 19 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter
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
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 45 1 6 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 1 4 411 1 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 0 1 0 0,1 -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 0 1 1 Page 1 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter
Precalculus Notes: Unit Polynomial Functions 86 E6: Graph g. 4 Domain: All reals,, Vertical Asymptotes:, 4 0 1 0, 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
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. 1 1 4 E1: Solve the equation. y y y 4 Factor to ind LCD: Multiply by LCD: y y y 1 1 4 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: 1 1 4 4 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. 4 8 5 4 6 16 151 Solve the equation by actoring: 6 10 4 0 4, 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
Precalculus Notes: Unit Polynomial Functions 4 E: Find the solution set o the equation 710 5. 4 5 5 5 Factor to ind LCD: Multiply by LCD: 5 5 Solve the resulting equation: 5 LCD = 4 5 140 68 815 0 5 0,5 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
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
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: 5 8 0 5 8 0 5,,8 + + Solution: 5,,8 Note: I the inequality was 5 4 8 0, 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. 1 1 1 1 1 Rewrite the inequality: 0 0 0 1 1 1 10 1 0 0 Critical Values: 1 1 + + Sign Chart: 1 Solution:,, 1 Page 6 o 7 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter
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:,.414 0.414,0.5 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. 58 58 116 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 10.975 inches. 1 You Try: Solve the inequality 0. 44 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