College algebra
posted by samantha on .
Find the equation with xintercepts (4,0),(3,0) and (2,0) and yintercept (0,12)

y=ax^3+bx^2+cx+d=a*(xx1)*(xx2)*(xx3)
where is x1,x2,x3 coordinates of
xintercepts
In this case x1=4, x2=3 and x4=2
y=a*(x4)*(x+3)*(x2)
=a*(x^24x+3x12)(x2)
=a*(x^2x12)*(x2)
For x=0
y=a*(12)*(2)=24*a= 12 Divide dwith 24
a=12/24=1/2
y=(1/2)*(x4)(x+3)(x2)
=(1/2)*(x^2x12)(x2)
=(1/2)*(x^3x^212x2x^2+2x+24)
=(1/2)*(x^33x^210x+24)
OR
y=(1/2)*x^3(3/2)*x^25*x+12