College algebra

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Find the equation with x-intercepts (4,0),(-3,0) and (2,0) and y-intercept (0,12)

  • College algebra -

    y=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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