Find an equation of the cubic polynomial f(x) = ax3 + bx2 + cx + d that passes through the given points.

Question

P(0,-4) Q(1,-8) R(-1,-8) S(2,-2)

Ive tried this every which way and i cant seem to get any of these..i know your supposed to plug it in and all but its just not working for me. can someone help me and write out the steps for me pleasee!

Answer 1

sub in each of the points to get

-4 = 0+0+0+d , so we know d = -4
-8 = a+b+c - 4 ---> a+b+c = -4
-8 = -a + b -c-4 ---> -a + b-c = -4
let's add these last two : 2b = -8 or b = -4

last point:
-2 = 8a +4b + 2c - 4
8a -16 + 2c - 4 = 2
4a -8 + c = 1 or c = -4a+9

sub into a+b+c=-4
a -4 + (-4a + 9) = -4
-3a = -9
a=3 and finally c = -12+9 = -3

so f(x) = 3x^3 -4x^2 - 3x - 4

I will leave it up to you to sub in the given points, I think they all work

BTW, the above is not the only way to do it, I sort of just let it flow.

Answer 2

Solve the system:
a*0^3+b*0^2+c*0+d=-4
a*1^3+b*1^2+c*1+d=-8
a*(-1)^3+b+(-1)^2+c*(-1)+d=-8
a*2^3+b*2^2+c*2+d=-2

d=-4
a+b+c+d=-8
-a+b-c+d=-8
8a+4b+2c+d=-2

Add the 2nd and the 3rd: 2b+2d=-16

d=-4
b=-4
a+c=0
8a+2c=18

Subtract from the 4th 2*3rd: 6a=18

a=3, b=-4, c=-3, d=-4