Unit 4 Polynomials Equations and Inequalities

Unit 4Polynomials Equations and Inequalities 4.1 Solving polynomial equations 4.2 Linear Inequalities 4.3 Polynomial Inequalities 4.4 Polynomial Rates of change

4.1 State the equation,zeroes, and degreeof this function.(0,2) is a point on the graph.

4.1 Equation: f(x)=-1/6(x+4)(x+1)(x-3)(x+4)(x+1)(x-3)2=a(0+4)(0+1)(0-3)2=a(4)(1)(-3)2=-12aa=2/-12a=-1/6 Zeroes : x= -4, x= -1, x= 3 Degree : 3

4.1 Determine the roots F(x)=x3 -8x2 -3x +90

4.1 f(x)=x3 -8x2 -3x +90 Sub in -3 f(-3)= (-3)3-8(-3)2 -3(-3) +90 f(-3)= 0 (x+3) is a factor x2 -11x +30 x2-11x +30 x+3 x2 -6x -5x +30 x(x-6) -5(x-6) Roots: x=-3, x=5, x=6 (x-6) (x-5)

4.1 -Application Determine algebraically where the cubic polynomial function that has zeros at 2, 3, and -5 and passes through the point (4,36) has a value of 120.

4.1- Application 36=a(4-2)(4-3)(4+5) It has a value of 12036=a(2)(1)(9) at x= -3, -2, -536=a(18)a=2 F(x)= 2(x-2)(x-3)(x+5) = 2(x2 -5x +6)(x+5) = 2(x3-19x +30) 120=2(x3 -19x +30) 60=x3-19x +30 0= x3-19x -30

4.2 Solve this inequality and state your solution using interval notation and a number line 2x<3x+6 4+2x 2

4.2 2x<3x+64+2x 2 4x<3x+6 8+4x x< 6 x+8 x<6 6 x+8 6-8 x -2 x xE [-2,6)

4.2 The New Canadian cell phone company charges $20 a month for service and $0.02 per minute of talking time. The My Mobile company charges $15 dollars a month for service and $0.03 per minute of talking time. a) Write expressions for the total bill of each companyb) Set up an inequality that can be used to determine for what amount of time(in minutes) my mobile is the better planc) Solve your inequality

4.2 N(t)=20 + 0.02t M(t)=15 + 0.03t 20 + 0.02t > 15 + 0.03t 20 + 0.02t > 15 + 0.03t 0.02t – 0.03t > 15-20 -0.01t > -5 t < 500

4.3 Determine an expression for f(x) in which f(x) is a quintic function, f(x) >0 , when -5<x<8 , f(x)< 0. When x is 7 there is a point of inflection. f(5) =45.

4.3 f(x) =a(x+5)(x-8)(x-7)3 f(5) =a(5+5)(5-8)(5-7)3 45 =a(10)(-3)(-2)3 45 =a(-30)(-8) 45 = 240a 240 240 9/50 = a

4.3 Solve the inequality x3-2x2+5x+20 > 2x2+ 14x-16

4.3 x3-2x2+5x+20 > 2x2+ 14x-16 x3-4x2-9x+36 > 0 x2(x-4)-9(x-4) > 0 (x-4)(x-3)(x+3) >0 (x-4)(x-3)(x+3)=0 -3,3,4 X<-3, -3<x<3, 3<x<4 , x>4

4.3 X3-2x2+ 5x+20 > 2x2+14x-16 when -3<x< 3 or x>4

4.4 For each of the following functions, calculate the average rate of change on the interval xE [2,5]a) f(x) = 3x + 1d) d(x) = -x² + 7f) v(x) = 9 For each of the functions in the question above, estimate the instantaneous rate of change at x=3.

4.4 A construction works drops a bolt while working on a high rise building 320 m above the ground. After t seconds, the bolt's height above the ground is s metres, where s(t) = 320 - 5t², 0 ≤ t ≤ 8. a) Find the average velocity for the interval 3 ≤ t ≤ 8.b) Find the velocity at t =2.

4.4 For each of the following functions, determine the average rate of change in f(x) from x=2 to x=7, and estimate the instantaneous rate of change at x=5. a) h(x) = (x-3)(2x+1)

4.4 The height in metres of a projectile launched from the top of a building is given by h(t) = -5t² + 20T +15 where t is the time in seconds since it was launched. a) How high was the projectile at the moment of launch?b) At what time does the projectile hit the round?c) What is the average rate of change in height from the time the object was launched until the time it hit the ground.

Review Questions Test book Page 240 #1ad, 3, 4, 5, 6bd, 7bc, 8cd, 9, 10a, 11, 14c, 15, 16, 17

Unit 4 Polynomials Equations and Inequalities