Write a cubic function that passes through (0,0),(-2,0),(1,0) and (-1,6)

Question

I can't figure this out at all, have tried experimenting different a values in a graphing calc but it has gone nowhere. Google doesn't seem to have any answers on finding cubic functions from 4 points that contain (0,0) as well.

Answer 1

The first three set of coordinates have Y = 0; therefore, the X coordinates are zeros of the function.

You can then write a(X-0) (X +2) (X - 1) = Y

This expands to a(X³ + X² -2X) = Y

When X = -1,Y = 6 Put these values into the last equation and solve for a. a = 3

The final equation is 3X³ + 3X² -6X = Y

Verify by entering the X values in the parentheses and solve; you should get the corresponding Y values.

Answer 2

If you are using a graphing calculator (like a TI84+ CE), first clear your y = equation (if there is one in there). Now enter the (x,y) points in the Stat function key (L₁ L₂ columns for x and y respectively). Once entered, you get out, then go back in but instead of edit, you would scroll to calc. then select cubic regression (option 6 on mine), scroll down to calculate, enter and it should give you all the coefficients in descending degree format (std format). I hope this helps. Let me know if you have any questions!