Review of Intermediate Algebra Content

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1 Review of Intermediate Algebra Content

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3 Table of Contents Page Factoring GCF and Trinomials of the Form + b + c... Factoring Trinomials of the Form a + b + c... Factoring Perfect Square Trinomials... 6 Factoring the Difference of Two Squares... 9 Factoring the Sum and Difference of Two Cubes... Factoring Strategy... Quadratic Equations... 8 Applications of Quadratic Equations... 0 Rational Epressions... Comple Fractions... 8 Rational Equations... 0 Applications of Rational Equations... Linear Equations... 0 Functions... 6 Algebra and Composition of Functions... 5 Eponents... 5 Radicals Rational Eponents... 6 Radical Equations Comple Numbers Solving Quadratic Equations... 7 Solving Quadratic and Rational Inequalities... 8 Equations That Can Be Written in Quadratic Form Graphing a Quadratic Function of the Form f ( ) a b c( a 0) = Pythagorean Theorem... 9 Eponential and Logarithmic Functions Vocabulary Used in Application Problems...00 Review of Intermediate Algebra...0 i Revised: May 0

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5 FACTORING FACTORING OUT THE GREATEST COMMON FACTOR. Identify the TERMS of the polynomial.. Factor each term to its prime factors.. Look for common factors in all terms.. Factor these factors out. 5. Check by multiplying. Eamples: Factor y Terms: 6, 8, and y + y y Factor into primes + y y Look for common factors Answer: ( y ) Check: ( ) ( y ) ( ) + Factor out the common factor + Check by multiplying y FACTORING TRINOMIALS OF THE FORM: + b + c To factor , look for two numbers whose product = 6, and whose sum = 5. Since the numbers and work for both, we will use them. = 6 + = 5 So, = ( + )( + ) Answer: ( + )( + ) Check by multiplying: ( )( ) List factors of 6: 6 Choose the pair which 6 adds to 5 = 6 + = =

6 FACTORING FACTORING OUT GCF AND FACTORING TRINOMIALS OF THE FORM: +b+c Problems Answers Factor ( 5 + 6). 7 5 y y y + 7y ( y + y ). 6 ( 9)( + 7). + ( 6)( ) 0 8

7 FACTORING FACTORING TRINOMIALS OF THE FORM: Eample: Factor STEP. Factor out the GCF ( ) a +b+c NOTES GCF is.. Identify the a, b, and c a = ; b= 7; and c= for the epression Find the product of ac ac = = 6. List all possible factor pairs that equal ac and identify the pair whose sum is b = 7 + = 5 and 6 and and are the only factor pairs, however and 6 is the correct pair. 5. Replace the b term with the boed numbers You have replaced + 7 with Group the two pairs. ( + ) + ( 6+ ) 7. Factor the GCF out of each pair. ( + ) + ( + ) The resulting common factor +. is ( ) 8. Factor out ( + ) ( )( ) + + These are the two factors of Recall the original GCF from Step. ( ) = ( + )( + ) There are factors of Answer: ( +)( +) 0. Check by multiplying. ( )( ) + + = 6 + +

8 FACTORING FACTORING TRINOMIALS OF THE FORM: Eample: Factor a a a +b+c Hint: Recall the rule of signs: When ac is negative, the factor pairs have opposite signs. STEP NOTES. Factor out the GCF. None. Identify the a, b, and c a = ; b= ; c=. Find the product of ac ac = ( ) =. List all possible factor pairs that equal ac and identify the pair whose sum is b. + ( ) + + ( ) = ( + ) + ( + ) =+ + ( 6) + + ( 6) = ( + 6) = + ( + 6) =+ + ( ) + + ( ) = ( + ) + ( + ) =+ Include all possible factor combinations, including the +/ combinations since ac is negative. 5. Replace the b term with the boed numbers. = + Note: a= a 6a a a a a 6a 6. Group the two pairs. = a + a 6a Group by pairs. The resulting common factor a + is ( ) 7. Factor the GCF out of each pair. ( ) ( ) = a a+ a+ 8. Factor out ( a + ) ( a )( a ) = + These are the factors. Answer: ( a+ )( a ) 9. Check to verify. ( )( ) a+ a = a a

9 FACTORING FACTORING TRINOMIAL PROBLEMS OF THE FORM: a + b + c Problems Answers Factor. ( a+ )( a ) 6a 5a ( a )( a+ ) 0a 5a 0. + ( )( 5 6) ( + 7)( 5+ ) 5

10 FACTORING FACTORING PERFECT SQUARE TRINOMIALS Eample: Factor STEP NOTES. Is there a GCF? No. Determine if the trinomial is a Perfect Square. Is the first term a perfect square? Yes ( ) = Is the third term a perfect square? +8 Yes 9 9 = ( 9) = 8 Is the second term twice the product of the square roots of the first and third terms? +8 Yes ( ) 9 = 8. To factor a Perfect Square Trinomial: Find the square root of the first term. = Identify the sign of the second term + Find the square root of the third term. 9 8 = 9 Write the factored form. Answer: ( +9 ) ( )( ) ( ) = = + 9 Check your work. ( ) ( ) =

11 Eample: Factor 5 + 0y + y STEP NOTES. Is there a GCF? No. Is the trinomial a Perfect Square Trinomial? Is the first term a perfect square? 5 Yes, = 5 = ( ) Is the third term a perfect square? y Yes, y y y = y = ( ) Is the second term twice the product of the square roots of the first and third terms? 0y Yes, ( ) 5 y = 0y. Factor the trinomial. Square root of the first term 5 = 5 Sign of the second term + 0y + Square root of the third term y = y Factored form. ( )( ) ( ) = = 5 + y y y y y Answer: ( 5+ y) Check your work. ( ) ( ) 5+ y 5+ y = 5 + 0y+ y 7

12 FACTORING FACTORING PERFECT SQUARE TRINOMIALS Problems Answers Factor ( ) +. y 8y 9 + ( ) y y 5y + ( ) 5y. 9 0y 5y + + ( ) + 5y 8

13 FACTORING FACTORING THE DIFFERENCE OF TWO SQUARES Eample: Factor: 6 STEP NOTES. Is there a GCF? No. Determine if the binomial is the Difference of Two Squares. Can the binomial be written as a 6, yes difference? Is the first term a perfect square? Yes, = Is the second term a perfect 6 Yes, 6 6 = 6 square?. To factor the Difference of Two Squares Find the square root of the first term Find the square root of the second term Write the factored form. Answer: ( + 6)( 6) = 6 6 = 6 ( )( ) Check your work. ( )( ) 6 = The factors are the product of the sum (+) and difference ( ) of the terms square roots = 6 Eample: Factor 6a b STEP NOTES. Is there a GCF? No. Is the binomial the Difference of Two Squares? Can the binomial be written as a 6a b Yes difference? Is the first term a perfect square? 6a Yes, 6a 6a= 6a Is the second term a perfect b Yes, b b= b square?. To factor the Difference of Two Squares Find the square root of the first 6a 6a = 6a term Find the square root of the b b = b second term Write the factored form. 6a b = Answer: ( 6a+ b)( 6a b) ( 6a+ b)( 6a b) Check your work. ( )( ) 6a+ b 6a b = 6a b The factors are the sum (+) and difference ( ) of the terms square roots. 9

14 FACTORING FACTORING DIFFERENCE OF TWO SQUARES Problems Answers Factor. 9 9 y ( y+ 7)( y 7). 5 6 ( 5 )( 5+ ). ( y)( + y) 9y. 8 6 a ( 9a + )( a+ )( a ) 0

15 FACTORING FACTORING THE SUM OF TWO CUBES Eample: Factor y 5 STEP. Is there a GCF? y = ( 8 + 5y ) NOTES Yes, is the GCF for both terms.. Determine if the binomial is the Sum or Difference of Two Cubes y Remember you are only looking at 8 + 5y ; Sum. Is the first term a perfect cube? 8 Yes, ( ) = = 8 Is the second term a perfect cube? + 5y Yes, 5y 5y 5y = 5y = 5y ( ). To factor the Sum of Two Cubes, use ( )( ) a + b y = a + by a aby + b y where a and b are constants. To factor 8 + 5y : Find the cube root of the first term. 8 = Find the cube root of the second term. 5y 5y = 5y Using the formula above, write the factored form. ( )( ) 8 + 5y = + 5y 0y + 5y Note the sign of the 0y is opposite that of the + 5y. Recall, you factored out the GCF in the final factored form y = ( + 5y)( 0y + 5y ) Answer: ( + 5 y)( 0y + 5 y ) from the original binomial. You must now include it

16 FACTORING FACTORING THE DIFFERENCE OF TWO CUBES Eample: Factor 6 7 y STEP NOTES. Is there a GCF? 6 7 y No. Determine if the binomial is the Sum or Difference of Two Cubes. 6 7 y Difference Is the first term a perfect cube? 6 Yes, ( ) = = 6 Is the second term a perfect cube? 7y Yes, ( ) y y y = y = 7y. To factor the Difference of Two Cubes, use ( )( ) a b y = a by a + aby + b y Where a and b are constants. To factor 6 7 y : Find the cube root of the first term. 6 = Find the cube root of the second term. y 7y = y Using the formula above, write the factored form. ( )( ) 6 7 y = y 6 + y + 9y Note the sign of y is opposite that of the 7y.

17 FACTORING FACTORING THE SUM and DIFFERENCE OF TWO CUBES Problems Answers Factor ( + )( 9 6+ ). 6a 5b ( a 5b)( 6a + 0ab + 5b ). y y y ( y )( y + y+ ) 5

18 FACTORING FACTORING STRATEGY. Is there a common factor? If so, factor out the GCF, or the opposite of the GCF so that the leading coefficient is positive.. How many terms does the polynomial have? If it has two terms, look for the following problem types: a. The difference of two squares b. The sum of two cubes c. The difference of two cubes If it has three terms, look for the following problem types: a. A perfect-square trinomial b. If the trinomial is not a perfect square, use the grouping method. If it has four or more terms, try to factor by grouping.. Can any factors be factored further? If so, factor them completely.. Does the factorization check? Check by multiplying.

19 FACTORING STRATEGY Always check for Greatest Common Factor first! Then continue: Terms Terms Terms Difference of Two Squares ( ) ( ) + ( )( ) Sum and Difference of Two Cubes 8= + + ( )( ) ( )( ) + 8= + + Perfect Square Trinomials ( ) + ( )( 5) + ( 5) ( + 5)( + 5) ( + 5) Form: + b + c 6+ 5 List factors of Choose the pair which adds to 6 5+ = ( 5) ( 5) ( 5)( ) Form: a + b + c = 0 List factors of Choose the pair which add to = 7 Substitute this pair in for the middle term Factor out the GCF of each pair of terms ( 5 + ) ( 5 + ) ( 5+ )( ) Factor by Grouping ( + ) ( + ) 7 6 ( + )( 7 6) 5

20 FACTORING Problems Factor a p 6q

21 FACTORING Answers. ( + )( )( + ). ( )( + ). ( + )( )( + ). ( + )( ). ( + )( )( ). ( + )( ). ( + )( )( + ) 5. 5 ( + )( 5 ) 5. ( + )( )( 5) 6. ( + 9)( + )( ) 6. ( 5)( 5)( ) + 7. ( + )( 9 6+ ) 7. ( )( 5) + 8. ( a)( + a+ 6a ) 8. ( )( ) + 9. ( 5 p q)( 5 p + 0 pq + 6q ) 9. ( )( ) 0. 0( )( + + ) 0. ( 6 5)( ) +. ( + 5)( 0+ 5). ( )( ) +. ( )( + + 9) 7

22 QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS BY FACTORING. Write the equation in standard form: a + b + c = 0 or 0 = a + b + c.. Factor completely.. Use the zero-factor property to set each factor equal to 0.. Solve each resulting linear equation. 5. Check the results in the original equation. Eample: Solve 00 = 0 This equation is in standard from. ( )( ) = 0 Factor completely. 0 = = 0 Set each factor equal to 0. = 0 = 0 Solve each linear equation. Check: = 0 Check: = 0 00 = 0 00 = 0 ( ) 0 00 = 0 ( ) 0 00 = = = 0 0 = 0 True 0 = 0 True The solutions of 00 = 0 are 0 = and = 0. Eample: Solve 6 = 7+ This equation is in standard 6 7 = 0 form. ( + )( ) = 0 Factor completely. + = 0 = 0 Set each factor equal to 0. Solve each linear equation. = = Check: = Check: = 6 = 7+ 6 = = 7 + = = + 6 = + 9 = True 7 7 = True The solutions of 6 7 = + are = and =. 8

23 QUADRATIC EQUATIONS Problems Answers Solve = 0 = 7, = = 5 = =. 5 = t 5 t = 5 t =. 8y 6y = 0 y = 0 y = 5. ( 0) 8 + = = = = 0 = 5 7. ( 6) = + = 0, =, = 9

24 APPLICATION OF QUADRATIC EQUATIONS Eample: Sharon and Diane went to the deli department at a local grocery store. Sharon arrived first and took a number to reserve her turn for service. Diane took the net number. If the product of their two numbers was 0, what were their numbers? Define the variable. Write and equation ( ) Solve the equation n = Sharon s number n + = Diane s number n n+ = 0 n + n= 0 n + n 0 = 0 ( n+ )( n 0) = 0 n = n = 0 - is an etraneous solution Sharon s number: n = 0, Diane s number: n + = Check in the original problem. The answers are: Sharon took number 0 and Diane took number. Eample: Anita s hat blows off her head at the top of an amusement park ride. The height h, in feet, of the hat t seconds after it blew off her head is given by h= 6t + 6t+. After how many seconds will the hat hit the ground? Define the variables h=height of the hat (in feet) t =time (in seconds) Use the given equation h= 6t + 6t+. Substitute h = 0 because the height of the hat is 0 feet when it hits the ground. Solve the Equation 0 = 6t + 6t+ 0 = 6( t t ) 0 = 6( t )( t + ) t = 0 t+ = 0 t = t = is an etraneous solution Check in the original problem. The answer is: It took seconds for the hat to hit the ground. 0

25 APPLICATION OF QUADRATIC EQUATIONS Problems Answers Solve. Bob has two routes he could take to get from his house to the mall. Bob s House a) 8 miles b) 6 miles 7 Mall 5 One route is direct and along a country road for 7 miles. The other route is miles south and 5 miles due east to the mall. a) Find the distance. b) Determine how many miles he saves driving the direct route.. Phil teaches in two different rooms. The room numbers are consecutive even integers whose product is 68. Find the two room numbers. and. The area of a garden is 95ft. The length is feet more than twice the width. Find the dimensions of the garden. 5 ft. and ft.. Suppose that a cannon ball is shot upward with an initial speed of 80 ft./second. The height h in feet of the cannon ball t seconds after being shot is given by h= 6t + 80t. After how many seconds will it hit the ground? 5 seconds

26 RATIONAL EXPRESSIONS SIMPLIFYING RATIONAL EXPRESSIONS Eample: Simplify 6 To simplify, factor numerator and denominator completely and cancel common factors. Recall =. 6 ( + )( ) ( ) + = = Eample: Simplify Recall: Step of factoring strategy: The leading coefficient must be positive. Continue by factoring the numerator and denominator completely. Cancel common factors ( ) = = = ( )( ) ( + )( + 5) ( ) + 5 OPERATIONS WITH RATIONAL EXPRESSIONS To add, subtract, multiply or divide rational epressions, use the following rules for adding, subtracting, multiplying or dividing fractions. Recall:. Multiply two fractions by multiplying the numerators and multiplying the denominators. Use cancellation, if necessary, to write the product in lowest terms. Eample: Simplify ( + )( ) ( + ) ( + )( ) = = + +

27 RATIONAL EXPRESSIONS. Divide two fractions by inverting the second fraction (divisor) and multiplying. Eample: Simplify a a a + a a+ a a a+ a a+ = a a+ a a a+ ( a+ )( a ) a + ( a ) ( a )( a ) ( a + ) ( a ) = =. To add (or subtract) like fractions, add the numerators only and place your result over the common denominator. Eample: Simplify = = To add (or subtract) unlike fractions, find the lowest common denominator, rewrite fractions with the common denominator, and add (or subtract) numerators, placing the answer over the common denominator. Eample: Simplify + + ( )( ) ( ) ( )( ) + = = + ( + )( ) ( + )( ) + = = ( + )( ) ( + )( ) Lowest common denominator is +. which is ( )( ) IMPORTANT NOTE: All answers must be simplified, i.e. reduced to lowest terms.

28 RATIONAL EXPRESSIONS Eamples:. Multiply: + + ( + )( ) = + ( + )( ) ( + )( ) = ( )( ) ( ). Multiply: 5 a a a 5 a = a a a a ( )( ) ( ) a 5 a a + a+ a = ( a + a+ ) a = a + ( + )( ) a a a a a a + + = a +. Divide: = ( + )( ) + 7 = ( ) = ( )( ) ( )( ). Subtract like fractions: ( ) ( + ) + 5 = = ( + ) + 9 = + = = ( + ) ( + )

29 RATIONAL EXPRESSIONS 5. Subtract unlike fractions: 5 y 7 5 y 7 = y y y y ( ) = 5 y 7 + y y = 5 + y 7 y 6. Subtract unlike fractions: y+ y+ y+ y+ = ( )( ) ( )( ) y y y y y y y y ( )( ) ( )( )( ) ( y+ )( y+ ) ( y+ )( y ) ( y 6)( y )( y+ ) ( ) y + y+ y + y ( y 6)( y )( y+ ) ( )( ) ( )( )( ) y+ y+ y+ y = y 6 y y+ y 6 y+ y = = y + y+ y y+ y + y+ = = ( y 6)( y )( y+ ) ( y 6)( y )( y+ ) 5

30 RATIONAL EXPRESSIONS Problems Answers Perform any operations. Simplify ( ) ( + ) y + y y y ( )( ) ( 5)( + ) ( + )( ) ( + ) + ( ) ( ) ( + ) ( )

31 RATIONAL EXPRESSIONS Problems Answers ( + ) ( + )( ). + 0 ( )( ) ( + )( ) y y + + 6( + y) ( )( + ) + + Note: To simplify rational epressions, factor the numerator and the denominator completely. Then cancel common factors to write the rational epression in lowest terms. 7

32 COMPLEX FRACTIONS SIMPLIFYING COMPLEX FRACTIONS. If needed, add or subtract in the numerator and/or denominator so that the numerator is a single fraction and the denominator is a single fraction.. Perform the indicated division by multiplying the numerator of the comple fraction by the reciprocal of the denominator.. Simplify the result, if possible. Eample: Simplify + + = + The LCD of the numerator of the comple fraction is. The LCD of the denominator of the comple fraction is. = + = + Simplify the numerator. Simplify the denominator. + = + = = ( + ) ( ) Change division to multiplication by the reciprocal. Cancel common factors. 8

33 COMPLEX FRACTIONS Problems Answers Simplify y y + 6 ( y ) y( y+ ) 5. 5 y y 5 y y+ 5 or y y. 8 b + 5 b 0 bb+0 ( ) 9

34 STEPS TO SOLVE RATIONAL EQUATIONS. Determine which numbers cannot be solutions of the equation.. Multiply all terms in the equation by the LCD of all rational epressions in the equation. This clears the equation of fractions.. Solve the resulting equation. RATIONAL EQUATIONS. Check all possible solutions in the original equation. Eample: Solve y + = y y STEP. y 0 y. y ( y ) + ( y ) = ( y ) y y. ( ) + ( y )( y ) = ( ) + y y+ = y y = 0 ( y )( y ) + = 0 y = 0 y+ = 0 y = y = 6 NOTES The denominator equals 0 when y =. Therefore, y = cannot be a solution. Multiply all terms in the equation by the LCD of ( y ). Cancel common factors. This clears the equation of fractions. Solve the equation. Check y = and y = Check all possible solutions in the original equation. Check y = Check y = + = + = = + = + = + = = The solutions are y = and y =. 0

35 RATIONAL EQUATIONS Eample: Solve 5 y + + = y y STEP. y 0 y NOTES The denominator equals 0 when y =. Therefore, y = cannot be a solution.. 5 y + y y ( y )( ) + ( y ) = ( y ) Multiply all terms in the equation by the LCD of y. Cancel common factors.. ( y )( ) + 5= y+ This clears the equation of fractions. y 8+ 5= y+ y = y+ y = Solve the equation.. Check y = = = 0 0 Check all possible solutions in the original equation. Cannot divide by 0. See Step. y = cannot be a solution. No solution ( is etraneous)

36 RATIONAL EQUATIONS Eample: Solve 6 = STEP NOTES The denominator equals 0 when = and = 7. Therefore = and = 7 cannot be solutions.. ( + 7) = 6( + ) This eample is a proportion so we can use the cross product.. + 7= = 0 + = 0 = = ( )( ) Solve this equation.. Remember to Check. The solutions are = and =.

37 RATIONAL EQUATIONS Problems Answers Solve. = y y y = 5. + = + 8 =. y 6 = y+ y+ 7 y =, y =. y 5 = y 5 y 5 No solution (5 is etraneous) = 9 + = 0

38 APPLICATIONS OF RATIONAL EQUATIONS NUMBER PROBLEM Eample: If the same number is added to both the numerator and denominator of the fraction 7, the result is. Find the number. Let = the unknown number + = 7+ + = 7+ ( + ) = 7 ( + ) 6 + = = = 5 Define the variable. Form an equation. Solve the equation. Check: + 5= = 9 = = Check the result in the original word problem. Answer: The number is 5. State the conclusion.

39 APPLICATIONS OF RATIONAL EQUATIONS UNIFORM MOTION PROBLEM d d Distance = Rate Time d = r t r = t = t r Eample: Tammi bicycles at a rate of 6 mph faster than she walks. In the same time it takes her to bicycle 5 miles she can walk miles. How fast does she walk? Rate Time = Distance bicycling w w + 6 walking w w Define the variable: w = Tammi s walking rate Check: 5 = w+ 6 w ( w ) 5w= + 6 5w= w+ w = w = walking rate is mph ( w = ) bicycling rate is 0 mph ( w + 6 = 0) Form an equation which sets the times equal to each other. (bicycling time = walking time) Solve the equation. Check the result in the original word problem. Using d t =, show that the walking time equals the bicycling time r t = walking miles mph bicycling 5miles t = 0mph t = hour t = hour The times are the same. Answer: Tammi s walking rate is mph. State the conclustion. 5

40 APPLICATIONS OF RATIONAL EQUATIONS WORK SHARED PROBLEM Work Completed = Work Rate Time Worked w= r t Work Completed by Unit + Work Completed by Unit = Job Completed Eample: It takes Charlie 6 hours to prepare the soccer field for a game. It takes Scotty 8 hours to prepare the same field. How long will it take if they work together? Work rate Time worked = Work completed Charlie t t 6 6 Scotty t t 8 8 Define the variable: t = time in hours that they worked together. Work Completed + Work Completed = Job Completed by Charlie by Scotty t t + = Form an equation. 6 8 t t + = 6 8 t + t = 7t = t = = hours 7 7 Solve the equation. Check: Using w= r t Charlie s work completed w = 6 7 Scotty s work completed w = 8 7 Check the result in the original word problem. w = of the job 7 + = job completed 7 7 w = of the job 7 Answer: It took them 7 hrs. working together. State the conclusion. 6

41 APPLICATIONS OF RATIONAL EQUATIONS SIMPLE INTEREST PROBLEM Interest = Principal Rate Time I I I = p r t = p = t rt pr Eample: An amount of money can be invested in a mutual fund which would yield $50 in one year. The same amount of money invested in a bond would yield only $50 in a year, because the interest paid is % less than that paid by the mutual fund. Find the rate of interest paid by each investment. Mutual Fund Bond Principal Rate Time = Interest 50 r 50 r 50 r.0 50 r.0 ( ) Define the variable: r = rate of interest paid by mutual fund r = Form an equation r.0 which sets the principals equal to each other. 50r 7 = 50r 00r = 7 Solve the equation. r =.05 =.5% r.0 =.05=.5% Check: Find the principal for each investment using I P = rt Check the result in the original word problem. Mutual Fund 50 P = = $0, 000 (.05)( ) Bond 50 P = = $0, 000 (.05)( ) The principals are the same. Answer: The interest rate for the mutual fund is.5% The interest rate for the bond is.5% State the conclusion 7

42 APPLICATIONS OF RATIONAL EQUATIONS Problems. If the same number is subtracted from both the numerator and the denominator of 0 9, the result is. Find the 5 number. Answers. If the numerator of 5 6 is increased by a certain number and the denominator is doubled, the result is. Find the number.. Melissa can drive 0 miles in the same time as it takes her to bike 0 miles. If she can drive 0 mph faster than she can bike, how fast can she bike? 5 mph. Mike can bicycle 6 miles in the same time that he can jog 9 miles. If he can bike 9 mph faster than he can jog, how fast can he bike? mph 8

43 APPLICATIONS OF RATIONAL EQUATIONS Problems 5. In year, an investor earned $60 interest on money he deposited in his savings account at a bank. He later learned that the money would have earned $50 had he invested in municipal bonds, because the bonds paid % more interest at the time. Find the rate he received in his savings account. Also, find the principal amount he put in the savings account. Answers % and $ A plane flies 780 miles downwind in the same amount of time as it takes to travel 580 upwind. If the plane can fly 0 mph in still air, find the velocity of the wind. 50 mph 7. Two bond funds pay interest rates that differ by %. Money invested for year in the first fund earns $0 interest. The same amount invested in the second fund earns $60. Find both the lower rate of interest and the principal amount of the investment. % and $ If the manager of a pet store takes 8 hours to clean all of the cages and it takes his assistant hours more than that to clean the same cages, how long will it take if they work together? 9 hours 9. If it takes a night security guard to make his rounds on campus hours and it takes his assistant hours, how long will it take them working together? 5 hours 9

44 LINEAR EQUATIONS The standard form of the equation of a line is written as a + by + c = 0 where a and b are not both 0. Definition of slope: Let L be a line passing through the points ( y ) and ( y ) of L is given by the formula:. Then the slope y y m = The slope-intercept form of the equation of a line is written as y = m + b where m is the slope and the point ( 0,b) is the y-intercept. The point-slope form of the equation of a line passing through (, y ) written as ( ) y y = m and having slope m is where m is the slope and (, y ) is the point. Parallel Property: Parallel lines have the same slope. Perpendicular Property: When two lines are perpendicular, their slopes are opposite reciprocals of one another. The product of their slopes is. 0

45 LINEAR EQUATIONS Eample: Find the equation of a line passing through ( 6,0 ) and ( 8, ) Find the slope of the line through the points. y y m =. 0 m = m = m = Use the point-slope form to find the equation of a line with Using ( 6,0) and m =, substitute into ( ) y y = m y 0 = 6 ( ) y 0 = m = and using one of the points. Answer: y = + 7 This is the equation of the line in slope-intercept form.

46 Eample: Find the equation of a line through ( ) Find the slope of the line. LINEAR EQUATIONS 9, and parallel to y =. y = y = m = This equation is in standard form. Write the equation in slope-intercept form. A line parallel to this line will have the same slope, m = Use m = and point (9, ) to write the equation of the line. ( ) y y = m y = 9 Answer: y = ( ) This is the point-slope form of a line. This is the slope-intercept form of a line. Eample: Find the equation of a line through (9, ) and perpendicular to y =. From the above eample, the line y = has A line perpendicular to this line will have a slope of Use m =. m = and point (9, ) to write the equation of the line. y = 9 ( ) m = (opposite reciprocal). Answer: y = +

47 LINEAR EQUATIONS Problems Answers. Put the following in slope-intercept form. a. 8 y = a. y = b. + y = b. y = +. Find the slope. a. (, ) and ( 7,9 ) a. 5 b. (,) and (, ) b. 5. Find the equation of the line through ( ) 5, and parallel to y = + y = +. Find the equation of the line, and perpendicular to y =. through ( ) y =

48 LINEAR EQUATIONS Find the equation of the line that satisfies the given conditions. Write the equation in both standard form and slope-intercept form. Answers Standard Form Slope-Intercept Form 5. slope = line passes through ( ), y = 5 5 y = + 6. slope = 5 line passes through ( ) 6 0,0 5+ 6y = 0 5 y = 6 7. slope = y-intercept y = y = 8. horizontal line through (, ) y = y = slope is undefined and passing through ( 5,6) = 5 cannot be written in the slope-intercept form. 0. slope = -intercept 5 + y = 5 5 y =. horizontal line through ( 5, ) y = y = 0

49 LINEAR EQUATIONS Find the equation of the line that satisfies the given conditions. Write the equation in both standard form and slope-intercept form. Answers Standard Form Slope-Intercept Form. vertical line passing through ( 5, ) = 5 Cannot be written in the slope-intercept form.. line passing through (, ) and ( ) 5, y = y = +. line passing through (, ) and ( 5, ) 5+ 8y = y = line passing through (, ) and (, 7) = Cannot be written in the slope-intercept form. 6. line parallel to + y = 6and passing through + y = 5 y = + 5 (, ) 7. line passing through ( 5, ) and parallel to y = 0 y = y = 0 8. line perpendicular to + 5y = and 5 y = 9 passing through (, 7 ) y = +

50 FUNCTIONS Recall that relation is a set of ordered pairs and that a function is a special type of relation. A function is a set of ordered pairs (a relation) in which to each first component, there corresponds eactly one second component. The set of first components is called the domain of the function and the set of second components is called the range of the function. Thus, the definition of function can be restated as: A function is a special type of relation. If each element of the domain corresponds to eactly one element of the range, then the relation is a function. Eamples: Determine if the following are functions. Domain Range y Domain Range y 7 5 This is a function, since for each first component, there is eactly one second component. This is not a function since for the first component, 7, there are two different second components, and 5. Eamples: Find the domain of the following functions.. 5 y = The denominator of a rational epression 0. Since + + 0,. The domain can be written as The domain is (, ) (, ). (, ) (, ).. y = Since this is a linear equation, there is an infinite number of values for, without restriction. The domain is any real number ( is any real number). The domain is (, ).. y = + The radicand of a square root must be 0. Since + 0,. The domain can be written as The domain is [, ). [, ). 6

51 FUNCTIONS Eamples: Find the range of the following functions.. y = Since is squared, the corresponding value for y must be 0. The range, 0 0,. The range is [ 0, ). y, can be written as [ ). y = Since this is a linear equation, there is an infinite number of values for y, without restriction. The range is any real number (y is any real number). The range is (, ).. y = + Since the square root yields an answer that is non-negative, y 0. The range, 0, 0,. The range is [ 0, ) y can be written as [ ) Vertical Line Tests: A graph in the plane represents a function if no vertical line intersects the graph at more than one point. Eamples: Use the Vertical Line Test to determine if the graph represents a function.. y This is a graph of a function. It passes the vertical line test.. y (,-) (,) This is not a graph of a function. For eample, the value of is assigned two different y values, and -. Since we will often work with sets of ordered pairs of the form (,y), it is helpful to define a function using the variables and y. Given a relation in and y, if to each value of in the domain there corresponds eactly one value of y in the range, then y is said to be a function of. 7

52 FUNCTIONS Notation: To denote that y is a function of, we write y = f ( ). The epression f ( ) is read f of. It does not mean f times. Since y and f ( ) are equal, they can be y =, or we can write f ( ) used interchangeably. This means we can write =. Evaluate: To evaluate or calculate a function, replace the in the function rule by the given value from the domain and then compute according to the rule. Eample:. Given: f ( ) = 6+ 5 Find: f (, ) f ( 0, ) f ( ) f f f ( ) ( ) = = 7 ( ) ( ) 0= 60+ 5= 5 ( ) ( ) = 6 + 5= g = Given: ( ) Find: g(, ) g( 0, ) g( ) ( ) ( ) ( ) g = 5+ 8= 6 g g ( ) ( ) ( ) 0 = = 8 ( ) ( ) ( ) = = 6 8

53 FUNCTIONS Problems Answers. Determine whether or not each relation defines a function. If no, eplain why not. a. Domain Range y 8 0 a. Yes, for each first component, there corresponds eactly one second component b. Domain Range y { } c. (, ), (, ), (,0) b. No, for each first component,, there corresponds two different second components, and 7. c. No, for the first component,, there corresponds two different second components, and 0. { } d. (, ),(, ),(,) d. Yes, for each first component there corresponds eactly one second component.. Determine the domain and range of each of the following: { } a. (, ),(, ),( 0,0 ),(,7) a. Domain: {,,0,} Range: { 0,,,7 } b. y b c. y c Domain:{ 5, 7, 9} Range: {,,6 } Domain: {,5,9 } Range: { 7,, } 9

54 FUNCTIONS Problems Answers. Given ( ) ( ), f = + and g = find the following: a. f ( ) a. 0 b. f ( ) b. c. g ( 0) c. d. g ( ) d. 5. Given k( ) = and h( ) 5 =, find the following: a. k ( ) a. 9 b. k ( 5) b. 5 c. h ( ) c. d. h( 7) d

55 ALGEBRA AND COMPOSITION OF FUNCTIONS OPERATIONS ON FUNCTIONS If the domains and ranges of functions f and g are subsets of the real numbers, then: The sum of f and g, denoted as f + g, is defined by ( f + g)( ) = f ( ) + g( ) The difference of f and g, denoted as f g, is defined by ( f g)( ) = f ( ) g( ) The product of f and g, denoted as f g, is defined by ( f g)( ) = f ( ) g( ) The quotient of f and g, denoted as f / g, is defined by f ( ) ( f / g)( ) = where g( ) 0 g( ) The domain of each of the above functions is the set of real numbers that are in the domain of both f and g. In the case of the quotient, there is the further restriction that ( ) 0 g. The composite function f g is defined by ( f g)( ) = f ( g( ) ) 5

56 ALGEBRA AND COMPOSITION OF FUNCTIONS Problems Answers Let f ( ) = 6 5 and g( ) = + Find: ( f + g)( ) = ( f g)( ) = ( f g)( ) = f g ( ) = f ( g( )) = ( f g)( ) = g( f ( )) = ( g f )( ) =

57 EXPONENTS LAWS OF EXPONENTS Definition: n a = a a a... a n times a = the base n = the eponent Properties: Product Rule: m n m n a a = a + Eample: + a a = a a a a a = a = a 5 times times Quotient Rule: a a m n m n = a Eample: 8 = = 8 6 n Power Rule: ( ) m a nm = Eample: ( ) a y = y = y Raising a product to a power: ( ) n n n ab = a b Raising a quotient to a power: n a a = b b n n Zero Power Rule: 0 a = Eample: ( ) 0 ( ) = ; = = = Eamples: Simplify ab a b c = a b c = a b c. ( )( ) 8 6. = = y y 8y. y ( y+ 5y ) = ( y )( y) + ( y )( 5y ) = 6 y + 5y = 6y + 5y 5

58 EXPONENTS NEGATIVE EXPONENTS Definitions: a m m a = a m a = a m a a b m n m b a m n m Note: The act of moving a factor from the numerator to the denominator (or from the denominator to the numerator) changes a negative eponent to a positive eponent. Eamples: Simplify... 0 ab 0 = = 0 0, 000 ab a b a a = = = 5 y yz = = 5 z b b Since b has a negative eponent, move it to the denominator to become b Only the factors with negative eponents are moved.. y y = y y y = 6 ( ) 6 ( ) y = Within the parenthesis, move the factors with negative eponents. Simplify within the parenthesis. Use the power rule for eponents. ( ) y = = Move all factors with negative eponents. y ( ) 6y = Simplify. 5

59 EXPONENTS RATIONAL EXPONENTS Eamples: Simplify. 5 5 = 5 = 5 +. = = = = 6 = 6 = 6. ( ) = = 5. ( ) ( ) ( ) ( ) a b = a b = ab y 5 = y 5 5 y 5 = y = y 5 55

60 EXPONENTS Problems Answers Simplify. a a a ( a ) a 5. ( ) 6. ( ) a a 7. ( 6y ) 8y 8. ( ab ) ab 9. y y / y 56

61 RADICALS Definition of Square Root of a The square root of the number a is b, a b b a =, if =. The radical symbol,, represents the principal or positive square root. The number or variable epression under the radical symbol is called the radicand. Eamples: Simplify 5 = 5 Check: ( 5) = 5 = Check: ( ) = = Check: ( ) = 6 6 Properties: is not a real number If " " a is a negative real number, a is not a real number.. Product Rule: a b = ab and ab = a b Eamples: Simplify = 6 8 = =. Quotient Rule: a = a and a = a b b b b Eamples: Simplify 9 9 = = 8 8 = = = = 57

62 RADICALS Problems Answers Use the product rule and the quotient rule to simplify each radical. All variables represent positive real numbers y y y. 8 y y m 0 m y 5 y 58

63 RADICALS Definition: n th root of a. The n th root of the number a is b, n a = b if For n a, n is the inde number and a is the radicand. Eamples: Simplify = Check: ( ) 8 = 8 n b = a. 6 = Check: ( ) = 6 The product and quotient rules for radicals apply for all inde numbers and larger (for all n ). a b = ab and ab = a b n n n n n n n n n a a a a = and = n n n b b b b Eamples: Simplify. 8 = 6 = Notice that 6 is a perfect cube. That is, 6 =.. 80 = 6 5 = 5 Notice that 6 is a perfect th. That is, 6 =.. ( + y) = ( + y) ( + y) 5 ( y) ( y) = y z = y y z = y z y Notice that 0 5 0, y, and z are all perfect 5 th s That is, ( ), y ( y), and z ( z ) = = =. = yz y 5 59

64 RADICALS Problems Answers Simplify. All variables represent positive real numbers y y y 7. ( + y) 0 ( + y) 7 ( + y) a b c a b c ac b b 7. 6 y 7 y 60

65 RADICALS ADDITION/SUBTRACTION OF RADICALS To have like radicals, the following must be true:. The radicals must have the same inde.. The radicands must be the same. To combine like radicals (i.e. add or subtract);. Perform any simplification within the terms.. Use the distributive property to combine terms that have like radicals. Eamples: Perform the operations. Simplify.. + = ( + ) 6 6 = 8. = ( ) = 5 8 These radicals are like radicals. Add or subtract the coefficients, using the distributive property.. + = ( + ) = = 8 9 = 8 = = = = = = = = ( ) = Create like radicals first. Then combine like radicals = 7 = ( ) = 6

66 RADICALS Problems Answers Perform the operations. Simplify y 00y y 9. 5y y 6y 0. 00y 8y y 6

67 RADICALS Problems Answers 5. 0a a 0 a. 8 6 y y r s t 0 8 r s t 6 rt. 8a a 6a y z y z y b b b a b c ab c a c ( 5)( 5) + 9 a. ( a)( a). ( 5+ )( 5 ) 6

68 RATIONAL EXPONENTS Definition of a Rational Eponent m n a = n a m n ( a ) m Eamples: Simplify 6 = 6 = ( ) 8 = 8 = ( ) ( ) 6 = 6 = 8 = 5 ( ) ( ) 7 = 7 = = 9 5 = = = ( ) ( ) ( ) ( ) abc = a b c = ab c 6 ( 8y) = = = y 6 6 ( 8y) 8 ( ) ( y ) ab = a b Use Power Rule of Eponents: ( ) n n n 6

69 RATIONAL EXPONENTS Problems Simplify Answers ( ) 6 8 ab ab 9. ( ) 0 b 8 5 b 0. ( ) 8 8a b a b. ( ) p q p q 7 65

70 RADICAL EQUATIONS To solve a Radical Equation: S. Isolate one radical epression on one side of the equation.. Raise both sides of the equation to the power that is the same as the inde of the radical.. Solve the resulting equation. If it still contains a radical, go back to step.. Check the results to eliminate etraneous solutions. Eample: Solve = 9 = 9 ( ) = ( 9) = = 0+ 8 ( )( ) 0 = 6 = = 6 Solve. Isolate the radical. The inde number is, so square both sides. Check = Check = 6 Check for etraneous solutions. = 9 ( ) Answer: = 9 6 ( ) 6 = = 9 9 = is a solution and = 6 is an etraneous solution. 66

71 RADICAL EQUATIONS Eample: Solve + 7 = + 7 = + Isolate one radical ( 7) ( ) + = + The inde number is, so square both sides. + 7= + + 6= = ( ) = The inde number is, so square both sides. 9 = Isolate the radical. Check = = = Answer: = 9 67

72 RADICAL EQUATIONS Problems Answers Solve.. 5 = =, =. + = =, =. 9 = = 0 ( 7 is etraneous) = 5 No solution is etraneous 5. = = = = 7 68

73 COMPLEX NUMBERS The imaginary number i is defined as i =. From the definition, it follows that i =. Eample: Simplify 9 9= 9 = 9 = i = i Eample: Simplify = 6 = 6 = i 6 = i 6 A comple number is any number that can be written in the form a + bi, where a and b are real numbers and i =. Eample: Write this comple number in a + bi form: = = 5+ 9i Eample: Simplify and write in a + bi form: 9 9 = 9 = 7i = 0+ 7i 69

74 COMPLEX NUMBERS Problems Write each imaginary number in terms of i.. 5 Answers 5i. i or i. 08 6i or 6 i. 6 8 i Write each comple number in the form a + bi i i or 5+ i i i i i 0 or i i 6 or 6 6 i 70

75 COMPLEX NUMBERS OPERATIONS WITH COMPLEX NUMBERS Eample: Perform this addition. Write the answer in a + bi form. ( + 5) + ( 5 ) ( + 5) + ( 5 ) = ( + 5i) + ( 5 i) ( 5) ( 5i i) = + + = 8+ i = i Combine like term. Eample: Perform this multiplication. Write the answer in a + bi form. ( 6 )( + ) ( 6 )( + ) = ( i)( + i) = + 6i 6i 8i = 0i 8i ( ) = 0i 8 = 0 0i = i Distribute Property Combine like terms. Recall i = Eample: Perform this division. Write the answer in a + bi form. 5 7i 5 5 i = 7i 7i i 5i = 7i 5i = 7 ( ) 5 = 0 i 7 Recall i = 7

76 COMPLEX NUMBERS COMPLEX CONJUGATES The comple numbers a + bi and a bi are called comple conjugates. Eample: Perform this division. Write the answer in a + bi form. 6+ i 5i 6+ i 6+ i + 5i = 5i 5i + 5i = ( 6+ i)( + 5i) ( 5i)( + 5i) Multiply the numerator and the denominator by the conjugate of the denominator i+ i+ 0i = 9 + 5i 5i 5i Distributive Property 8 + i+ 0i = 9 5i Combine like terms. ( ) ( ) 8 + i + 0 = 9 5 Recall i = + i = = + i a + bi form = + i 7 7 7

77 COMPLEX NUMBERS Problems Answers Perform the indicated operations. Write the answer in a + bi form.. ( 5i) ( 0 i) i. ( 7 6 ) + ( ) 6 + i. 5i( 8 6i) i. ( 7i)( 5 i) i 5. ( 7 )( + 6) + i i 6 0 i i i i + i i i i + i 0 5i 7

78 SOLVING QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS USING THE SQUARE ROOT PROPERTY Eample: Solve 8 = 0 = 8 0 = 8 Isolate the perfect square. = ± 8 Find the square root of both sides using the Square Root Property. =± 9 =± Remember to use ± sign when finding the square roots. = ± The solutions are = and =. Eample: Solve ( ) = 6 ( ) = 6 ( ) =± 6 The perfect square is already isolated. Find the square root of both sides using the Square Root Property. =± = = = 7 = =± needs to be solved with two separate equations. The solutions are = 7 and =. 7

79 SOLVING QUADRATIC EQUATIONS Eample: Solve + = = 6 0 = 6 Isolate the perfect square. =± 6 Remember to use the ± sign when = ± 8i The solutions are = 8 i and = 8i. finding the square roots. Recall that = i. Eample: Solve = = 0 6 = 9 9 = 6 9 = ± 6 = ± = ± 7i = ± 9 6 ( ) 7 The solutions are 7 7 = i and = i. 75

80 SOLVING QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS USING SQUARE ROOT PROPERTY Problems Answers Solve the equations. Be sure to get two solutions.. = 6 = ±. = 5 = ± 5. = 0 = ± 0. = = ± 5. = 9 = ± 6. = ± = 00 0 i 7. = = ± i 6 8. = 50 = ± 5i 76

81 SQUARE QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS USING SQUARE ROOT PROPERTY Problems Answers = = 7 and = 5 9. ( ) 6 + = = 5 and = 9 0. ( ) 9 = = 0 and = 0. ( ) 5 5. ( + ) = 8 = ± + = = ± i. ( ) = = ± i. ( ) 9 5. ( 8) = 60 = 8 ± 5 6. ( ) 0 + = = ± i 0 7. ( ) 8 = = ± i 77

82 SOLVING QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS USING FACTORING The Zero-Factor Property When the product of two real numbers is 0, at least one of them is 0. If a and b represent real numbers, and if ab = 0 then a = 0 or b = 0 Eample: Solve using factoring 7 = = Set equation equal to 0. ( )( ) + = 0 Factor completely. + = 0 or = 0 Use the Zero-Factor property to set each factor equal to 0. + = 0 = 0 = = = Solve the resulting linear equations. Check: = Check: = 7 0 = 7 = 0 ( ) 7( ) = 0 7 = = 0 + = = 0 + = 0 0= 0 Check the results in the original equation. The solutions are = and =. 78

83 SOLVING QUADRATIC EQUATIONS SOLVING QUADRATIC EQUATIONS WITH THE QUADRATIC FORMULA Eample: Solve 7 = = Set the equation equal to zero. a=, b= 7, and c = Compare this equation with the Standard Form a + b + c = 0 and identify the a, b and c values in this equation. b ± b ac = a ± = ( 7) ( 7) ( )( ) ( ) ( ) + 7 ± 9 8 = 8 7 ± ± 8 = = 8 8 7± 9 = 8 Substitute the values for a, b and c into the Quadratic Formula. b ± b ac = a = = = = = = Check: = Check: = 7 = 0 7 = 0 ( ) 7( ) = 0 7 = = = = = 0 0= 0 Check the results in the original equation. The solutions are = and =. 79

84 SOLVING QUADRATIC EQUATIONS The Discriminant: For a quadratic equation of the form a + b + c = 0 with real-number coefficients and a 0, the epression b ac is called the discriminant and can be used to determine the number and type of the solutions of the equation. Discriminant: b ac Number and type of solutions Positive 0... Negative.. Two different real numbers One repeated solution, a rational number Two different imaginary numbers that are comple conjugates 80

85 SOLVING QUADRATIC EQUATIONS Problems Answers Solve = 0 = 7 and =. = 0 = and =. = = and = = 0 = ± 6 5. = 5 = 0 and = = 6 = and = 7. 0 = = 6 and = = 0 = ± 8

86 SOLVING QUADRATIC EQUATIONS Problems Answers Solve = = ± = 0 = ± i. = = ± i. = 5 ± = = ± = 0 = 5 ±. = ± = = ± = 0 ± = = ± 8

87 SOLVING QUADRATIC AND RATIONAL INEQUALITIES TO SOLVE A QUADRATIC OR RATIONAL INEQUALITIES:. Write the inequality in standard form with 0 on the right side of the inequality.. Solve the related equation. The solutions are called critical numbers.. Consider any restrictions on the domain. For eample, in rational inequalities since the denominator cannot equal 0, set the denominator equal to 0 to determine the restrictions.. Locate the numbers from Steps and on a number line to create intervals. Determine if these numbers are included in the solution. 5. Test a number from each interval back in the original inequality. If it makes the inequality true, the interval is included in the solution. Eample: Solve the inequality 0 < < 0 < 0 0 Write the inequality in standard form. 0 0 = ( )( ) + 5 = 0 = = 5 Solve the related equation. (There are no restrictions on the domain)., 5,5 ( 5, ) Locate the numbers on a number line. Create the intervals. and 5 are not in the solution set. Eample continues on the net page. 8

88 SOLVING QUADRATIC AND RATIONAL INEQUALITIES Test a number from each interval into 0 <. test = test = 0 test = 6,,5 5 ( 5, ) Test a number from each interval in the original inequality. For interval,, test =. < 0 ( ) ( ) < 0 + < 0 6 < 0 False, this interval is not included in the solution set. For interval,5, test = 0. < 0 ( ) ( ) 0 0 < 0 0 < 0 True, this interval is included in the solution set. For interval ( 5, ), test = 6. < 0 ( ) ( ) 6 6 < 0 0 < 0 False, this interval is not included in the solution set. The solution is,5. 8

89 SOLVING QUADRATIC AND RATIONAL INEQUALITIES Eample: Solve the inequality = = = 0 ( ) ( ) = The inequality is in standard form. Solve the related equation. Consider restrictions on the domain. 5 Locate the numbers on a number line. (, ] [, 5) ( 5, ) Create the intervals. is in the solution set but 5 is not. For interval (, ], test = Test a number from each interval in the original inequality. True, this interval is included in the solution set. Eample continues on the net page. 85

90 SOLVING QUADRATIC AND RATIONAL INEQUALITIES For interval [, 5), test = False, this interval is not included in the solution set. For interval ( 5, ), test = True, this interval is included in the solution set. The solution is (, ] ( 5, ). 86

91 SOLVING QUADRATIC AND RATIONAL INEQUALITIES Solve the inequalities. Problem Answer. 6 6,. > 8 (, ) (, ). 5 < 0 + (, 5). + 0 [, ] (, ) 87

92 EQUATIONS THAT CAN BE WRITTEN IN QUADRATIC FORM SOLVING EQUATIONS USING U-SUBSTITUTION EXAMPLE: Solve for " " ( ) ( ) = 0 ( ) ( ) = 0 This equation is quadratic in form. u 5u+ 6= 0 Substitute " u " in the equation for + Let + = u. ( ) ( u )( u ) = 0 u = 0 u = 0 u = u = Factor and solve. + = + = = 0 = To solve the original equation for " ", substitute " + " in for " u " (Let u = + ). Check = 0 ( ) ( ) = 0 ( ) ( ) 5 ( ) = 0 Check = ( ) ( ) = 0 ( ) ( ) ( ) = 0 The solutions are = 0 and =. 88

93 EQUATIONS THAT CAN BE WRITTEN IN QUADRATIC FORM SOLVING EQUATIONS USING U-SUBSTITUTION Eample: Solve for " " + 7 8= = 0 This equation is quadratic in form = 0 ( ) u + 7u 8= 0 Substitute " u " in the equation for Let = u ( ) ( u )( u ) + 8 = 0 u = 0 u+ 8= 0 u = u = 8 Factor and solve = = 8 = = 8 = = To solve the original equation for " ", substitute in for " u ". Let u = ( ) Check = Check = = = 0 ( ) ( ) = 0 ( ) ( ) ( ) = 0 The solutions are = and =. 89

94 EQUATIONS THAT CAN BE WRITTEN IN QUADRATIC FORM Problems Answers Solve using U-Substitution.. ( ) ( ) = 0 = and = 0. ( ) ( ) = 0 = and = 5. + = 0 =± and =± i = 0 = and = = 0 = and = = 0 = 7,776 and = 90

95 GRAPHING A QUADRATIC FUNCTION OF THE FORM ( ) = f a b c ( a ) The graph of f ( ) = a + b + c is a parabola. Use the coefficients a, b, and c to find the following:. The parabola opens upward when a > 0. The parabola opens downward when a < 0.. The -coordinate of the verte of the parabola is verte is b b b f. The verte is, f a a a. b =. The y-coordinate of the a. The ais of symmetry is the vertical line passing through the verte.. The y-intercept is determined by the value of ( ) ( 0, f ( 0) ) or ( 0, c ). f when = 0 : the y-intercept is 5. The -intercepts (if any) are determined by the values of that make f ( ) = 0. The - intercept/s, if any, is (,0). To find them, solve the quadratic equation a + b + c = 0. 9

96 GRAPHING A QUADRATIC FUNCTION OF THE FORM f = a + b + c ( a ) f = 8+ 6 Eample: Graph ( ) ( ) 0 a=, b= 8, c= 6 Determine the values of a, b, and c.. a = ; parabola opens upwards The parabola opens upward ( 0) a >.. 8 = = f f ( ) ( ) ( ) ( ) ( ) = = 8+ 6 b a The verte is, f = (, ) b The -coordinate of the verte is =. a b The y-coordinate of the verte is f. a. = is the ais of symmetry. Ais of symmetry is the vertical line passing through the verte.. f ( 0) = 0 ( ) 80 ( ) + 6 f ( 0) = 6 The y-intercept is ( 06, ) 5. ( ) f = = 8+ 6 ( ) 0= + ( )( ) 0= = 0 = 0 = = Let = 0. The y-intercept is ( 0, ( 0) ) f. Let f ( ) = 0. The -intercept/s, if any, is (,0). The -intercepts are ( 0, ) and (, ) y Notes ais of symmetry: = y-intercept: (0,6) -intercepts: (,0) and (,0) 9

97 GRAPHING A QUADRATIC FUNCTION OF THE FORM f = a + b + c ( a ) ( ) 0 Problems Answers Use the 5-step process for graphing these quadratic functions.. ( ) f = + 6 y Step.Opens upward since > 0. Step. Verte is (, 8). Step. Ais of symmetry is =. 0, 6. Step. y-intercept is ( ) Step 5. The -intercepts are (,0) and ( ), 0.. ( ) f = 5 Step. Opens upward since > 0. y Step. Verte is (, 6). Step.Ais of symmetry is =. 0, 5. Step. y-intercept is ( ) Step 5. -intercepts are ( 7,0 ) and ( 5,0).. ( ) f = + Step. Opens downward since < 0. y Step. Verte is (,). Step. Ais of symmetry is =. Step. y-intercept is ( 0, ). Step 5. -intercepts are (,0) and (,0). 9

98 PYTHAGOREAN THEOREM In a right triangle the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the legs (the sides that form the right angle). If c is the hypotenuse and a and b are the lengths of the legs, this property can be stated as: c b c = a + b or c= a + b a a = c b or a= c b b = c a or b= c a Eamples: Find the length of the unknown side in the following right triangles.. Given a= and b= 5, find c. c = c =. Given b= and c=, find a. a = 69 a = 5 a = 5. Given c= 6 and a= 5, find b. b = 6 5 b = 9

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