Math-3 Lesson 4-6 Polynomial and Rational Inequalities

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1 Math-3 Lesson 4-6 Polynomial and Rational Inequalities

2 SM3 HANDOUT 4-6 Polynomial and Rational Inequalities Graph the general shape of the equation. y 4 1 Positive lead coefficient, even degree nd degree polynomial (has a maimum of -intercepts) Odd multiplicities (graph passes thru -ais at each zero.

3 In Lesson 4-4 We learned the Boundary Numbers Method of solving a quadratic inequality. 8 1 When solving quadratic equations, we first rearranged the equation to be in standard form. When solving quadratic inequalities, we first rearrange the inequality into standard form Find the boundary numbers solve the equation ( 14)( = 14, - ) - 14

4 ( 14)( ) = 14, - For two boundary numbers, the solution is either: 1) Between the boundary numbers or - 14 ) Outside of the boundary numbers - 14 Test a value to see if it is a solution. Zero is often the best number to test. 0 (0) 1(0) Is 0 a solution? (does = 0 make the inequality true?) The shaded part of the graph is the solution we must pick the option that shades the number 0. (, ] [14, )

5 Solve Single Variable Polynomial Inequalities For 3 rd degree and higher degree polynomials, there are often more than two zeroes. So we need to work harder to find the solution. Method 1: Sign (+/-) Table (Required for Math-1050) Method : Sign (+/-) Chart (Required for Math-1050 Method 3: Graphically (required for Math-3 Only)

6 Method 1: Sign (+/-) Table ( positive numbers) 1. Find the real zeroes of the polynomial equation ( 14)( ) =, 14. Identify the intervals (between the zeroes),, 14 [14, ) 3. Build a table to organize the information intervals Input ( 14)( + ) output Solution? (, ] 3 ( )( ) (+) yes [, 14] 0 ( )(+) ( ) no [14, ) 15 (+)(+) (+) yes 4. Write the Solution in interval notation =, [14, )

7 Method : Sign (+/-) Chart Find the real zeroes of the polynomial equation.. Divide the -ais into intervals using the zeroes Determine the sign (+/-) for each interval 0 ( 14)( ) 0 ( positive numbers) 0 ( ) 0 ( 14)( ) =, 14,, 14 [14, ) ( )( ) ( ) 4. Write the interval of -values that make the inequality true (positive intervals). ( )( ) ( ) (, ] [14, ) ( )( ) ( )

8 Method : Sign (+/-) Chart You can build the sign chart by replacing 0 with y then drawing a rough graph using the zeroes (and their multiplicities) and the end-behavior ( positive numbers) 0 ( ) 0 ( 14)( ) =, 14 y = 14 + = ,, 14 [14, ) The epression will be positive for regions where the graph is above the -ais (and negative below) =, [14, )

9 Method 3: Graphically y Graph on calculator Change to standard form inequality Change to standard form equation Replace 0 with y + + Determine -intercepts 14 Build function sign chart Solve ( numbers) 8 (, ] [14, )

10 Solve using the sign change method (show your work!!) 0 > ( + 5)( 6) 0 > (negative numbers) 0 ( ) Solution: -values that make the epression negative intervals Input (+5)(-6) output Solution? (, 5) 6 ( )( ) (+) no ( 5, 6) 0 (+)( ) ( ) yes (6, ) 7 (+)(+) (+) no 0 > ( + 5)( 6) = ( 5, 6)

11 Solve using the Sign Chart or table method (show your work!!) , 0 0 positive # s and zero 0 ( + )( ) Solution: -values that make the epression positive or zero intervals Input ( + )( ) output Solution? (, ) 3 ( )( ) (+) yes (, ) 0 (+)( ) ( ) no (, ) 3 (+)(+) (+) yes 0 4 =, U(, )

12 0 ( 1)( 1)( ) 0 ( positive numbers) 1. Find the real zeroes of the polynomial equation. 0 ( 1)( 1)( ) 1, 1,. Build Sign (+/-) Table intervals Input ( 1)( + 1)( ) output Solution? (, 1) ( )( )( ) ( ) no ( 1, 1) 0 ( )(+)( ) (+) yes (1, ) (, ) (+)(+)( ) ( ) no (+)(+)(+) (+) yes 3. Write the solution. = ( 1, 1) [, )

13 0 ( negative numbers) 0 ( ) 0 ( positive 0 ( ) Epect the following types of polynomais: numbers) 1. Nice Pattern these are the numbers in the bo Quadratic form factor using m-substitution 3. is a common factor factor

14 Solving Single Variable Rational Inequalities We will solve inequalities similar to the following: 3( 6)( ) ( )( 1) 0 ( positive numbers) 0 ( ) Solution: -values that make the rational epression positive 1) Zeroes of the numerator are -intercepts ( negative numbers) 0 ( ) Solution: -values that make the rational epression negative ) Zeroes of the denominator that do not disappear due to simplification are vertical asymptotes. 3) Zeroes of the denominator that disappear due to simplification are holes.

15 ( ) y 16 ( 4)( 4) ( ) (+) ( ) (+) 0 ( positive 0 ( ) Passes thru means opposite sign on each side of the zero. numbers) Solution: -values that make the rational 4 4 epression positive 1) Zero of the numerator: -intercept: (Passes thru at = ) ) Zero of the denominator: (Either vertical asymptote OR hole). No holes and VA at = -4, 4 interval Input ( )( + 4)( 4) output Solution? (, 4) ( 4, ) 5 0 ( )( )( ) ( )(+)( ) ( ) (+) no yes (, 4) (4, ) 3 5 (+)(+)( ) ( ) no (+)(+)(+) (+) yes = ( 4, ) [4, )

16 ( 4 1) y ( )( 1) 3( 6)( ) y ( )( 1) 0 0 and postiive # s 0 0, (+) (+) ( ) (+) ( ) (+) 6 1, 6 X-intercepts: (No holes), VA s are: = -, -1 Solution: =, 6, 1 [, ) interval Input ( + 6)( )( + )( + 1) output Solution? (, 6] 7 ( )( )( )( ) (+) yes [ 6, ) 3 (+)( )( )( ) ( ) no (, 1) 1.5 (+)( )(+)( ) (+) yes ( 1,] 0 (+)( )(+)(+) ( ) no [, ) 3 (+)(+)(+)(+) (+) yes

17 0 > ( + 7)( ) 0 > + ( ) ( + 7) 0 > + 1. X-intercepts: = -7. Vertical asymptote: 3. Hole: -7 - = = - interval Input ( + 7)( + ) output Solution? (, 7] 8 ( )( ) (+) no [ 7, ) 3 (+)( ) ( ) yes (, ) 0 (+)(+) (+) no (, ) 3 (+)(+) (+) no

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