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September 30, 2014

September 30, 2014

Posted by **samantha** on Tuesday, October 26, 2010 at 8:57pm.

- College algebra -
**Bosnian**, Wednesday, October 27, 2010 at 4:54amy=ax^3+bx^2+cx+d=a*(x-x1)*(x-x2)*(x-x3)

where is x1,x2,x3 coordinates of

x-intercepts

In this case x1=4, x2=-3 and x4=2

y=a*(x-4)*(x+3)*(x-2)

=a*(x^2-4x+3x-12)(x-2)

=a*(x^2-x-12)*(x-2)

For x=0

y=a*(-12)*(-2)=24*a= 12 Divide dwith 24

a=12/24=1/2

y=(1/2)*(x-4)(x+3)(x-2)

=(1/2)*(x^2-x-12)(x-2)

=(1/2)*(x^3-x^2-12x-2x^2+2x+24)

=(1/2)*(x^3-3x^2-10x+24)

OR

y=(1/2)*x^3-(3/2)*x^2-5*x+12

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