S

Math Project

Bader saeed hamad ALGhaithi Ali mubabrak aldhaheri

G:11.01

Task 1

x0 x1 x2

x -1 1 2 5

P(x) 10 -6 -17 82

P(x)= A+B(x-x0)+c(x-x0)(x-x1)+D(x-x0)(x-x1)(x-x2)

A=10

-6= A+B(x-x0)

-6= A+B(1-(-1))

-6= A+2B

-17= A+B(x-x0)+c(x-x0)(x-x1)

-17= A+B(2-(-1))+C(2-(-1))(2-1)

82= A+B(x-x0)+c(x-x0)(x-x1)+D(x-x0)(x-x1)(x-x2)

82= A+B(5-(-1))+C(5(-1))(5-1)+D(5-(-1))(5-1)(5-2)

A= 10

-6= A+2B

-6= 10+2B

-16= 2B

B= -8

-17= 10+(-24)+3C

-17=(-14)+3C

-3=3C

C=-1

82= A+6B+24C+72D

82= 10+(-48)+(-24)+72D

82= -62+72D

144= 72D

D= 2

P(x)= A+B(x-x0)+c(x-x0)(x-x1)+D(x-x0)(x-x1)(x-x2)

P(x)=10+(-8)(x-(-1))+(-1)(x-(-1))+(-1)(x-(-1))(x-1)+2((x-(-1))(x-1)(x-2)

2x3-5x2-10x+5

Task 2

A- Show that the 3 zeros of the polynomial found in task 1 are:

First zero lies between -2 and -1

Second zero lies between 0 and 1

Third zero lies between 3 and 4.

Find to the nearest tenth the third zero using the Bisection Method for Approximating Real Zeros.

   

Since f (-2) = -9 and f (-1) = -2 there is one zero between -1, -2.

 Since f (0) = -3 and f (1) = -4,there is zero between 0, 0.5,

 Since f (3) = 24 and f (4) = 65 there is zero between 2.5, 3

d) (2x+4) (x+4) – 2x2= 1000

2x2 + 8x + 4x + 16 – 2x2= 1000

12x + 16 = 1000

12x = 1000-16

12x = 984

X = 83

Task 4

A- Use a graphing program to graph the polynomial found in task 1

B- Make a PowerPoint to present your project and upload it on a wiki.