Math-3 Lesson 4-6 Polynomial and Rational Inequalities

Math-3 Lesson 4-6 Polynomial and Rational Inequalities

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.

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. 0 1 8 Find the boundary numbers solve the equation. 0 1 8 0 ( 14)( = 14, - ) - 14

0 1 8 0 ( 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) 8 0 8 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, )

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)

Method 1: Sign (+/-) Table 0 1 8 0 ( positive numbers) 1. Find the real zeroes of the polynomial equation. 0 1 8 0 ( 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, )

Method : Sign (+/-) Chart 0 0 1 8 1 8 1. Find the real zeroes of the polynomial equation.. Divide the -ais into intervals using the zeroes 14 3. 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, ) ( )( ) ( )

Method : Sign (+/-) Chart 0 0 1 1 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. 8 8 0 ( positive numbers) 0 ( ) 0 ( 14)( ) =, 14 y = 14 + = 1 8-14,, 14 [14, ) The epression will be positive for regions where the graph is above the -ais (and negative below) =, [14, )

Method 3: Graphically 8 0 0 y 1 1 8 1 8 1 8 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 0 1 0 ( numbers) 8 (, ] [14, )

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)

Solve using the Sign Chart or table method (show your work!!) 0 4 0 +, 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(, )

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, ) (, ) 1.5 3 (+)(+)( ) ( ) no (+)(+)(+) (+) yes 3. Write the solution. = ( 1, 1) [, )

0 ( negative numbers) 0 ( ) 0 ( positive 0 ( ) Epect the following types of polynomais: numbers) 1. Nice Pattern these are the numbers in the bo 0 0 3 4 10 4 9 8 3 0 6 16. Quadratic form factor using m-substitution 3. is a common factor factor

Solving Single Variable Rational Inequalities We will solve inequalities similar to the following: 3( 6)( ) 7 10 0 0 ( )( 1) 0 ( positive numbers) 0 ( ) Solution: -values that make the rational epression positive 1) Zeroes of the numerator are -intercepts 6 16 0 ( 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.

3 6 0 3( ) 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, )

0 3 + 1 36 + 3 + 3( 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

0 > + 10 8 4 ( + 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