Write a cubic function that goes through the points (1,0), (-1,0), (4,0), (2,-12)?

Question

can you please do it step by step. we have a test tomorrow on this and i still don't really understand how to do it. i missed the day of notes.

Answer 1

well, youll always want f(1) to be 0, f(-1) to be 0 and f ( 4) to be 0.

(x-1)(x+1)(x-4)=0

so that all those will be 0

sub 2 for x... (2-1)(2+1)(2-4)= -6

Multiply by 12(what you want)/(-6) (what you have)

12/-6=-2

so -2(x-1)(x+1)(x-4) should be your answer

Answer 2

Roots are:
(x-1)(x+1)(2x-8) = y

(x^2 - 1)(2x-8) = y

2x^3 - 8x^2 - 2x + 8 = y

Is the equation.

Proof:

2*2^3 - 8*2^2 - 2*2 + 8 = y

16 - 32 - 4 + 8 = y

-12 = y

I cheated and used guess and check methodology to realize (2x-8) and not (x-4) was needed when I saw (x - 4) came up with the - 6 = y. So just multiply the whole set of roots by 2 and that fixes it.

Answer 3

Since 1,-1, and 4 are the root of the eq

y = a(x-1)(x+1)(x-4)

use (2,-12) to find a
-12 = a*1*3*(-2)
a = 2

so the answer should be 2(x-1)(x+1)(x-4)

now multiply the factors out
y = 2x^3 - 8x^2 -2x + 8

Answer 4

let the function be
f(x)=a(x^3)+b(x^2)+c(x)+d
the function will pass through these points
solving the equations
you will get the answer by putting value of x
a+b+c+d=0
-a+b-c+d=0
64*a+16*b+4*c+d=0
8*a+4*b+2*c+d=-12
I would have solve it . But I really Dont have the time to Solve it

Answer 5

sorry i have no idea i hope you figure it out if you do let me know how