What is the estimated cubic function for given x and y-intercepts?

[ShawnPrend]

Homework Statement

X-intercepts: (-1.57,0) , (0.65, 0) , (2.83, 0)
Y-intercept: 11.33

Homework Equations

I've got to convert that information into an estimated cubic function.

The Attempt at a Solution

I tried subbing the y-intercept in; although that didn't work.

11.33 = K(1.57)(-0.65)(-2.83)
11.33 = K(2.888)
K = 3.923

That didn't provide the correct equation when subbed into

f(x)=K(x+1.57)(x-0.65)(x-2.83)

Can anyone help me on this?

 

[symbolipoint]

I checked your solution process and tried your results. No problem found. You only have four points to use and to check. They all work in your function which you found.

 

[HallsofIvy]

Why do you say "that didn't provide the correct equation"?

Certainly f(x)= 3.923(x+ 1.57)(x- 0.65)(x- 2.83) is 0 at x= -1.57, x= 0.65, and x= 2.83 and, as you calculated f(0)= 3.923(.157)(-0.65)(2.83)= 11.33. It's a cubic and it satisfies all the requirements.

 

[ShawnPrend]

It was my own fault; I mixed up a few things when I was graphing the equation on a graphing calculator.

Thanks for the help though.