I have to write a cubic function with these points (1,0)(3,0)(-2,0) (2,1)?

Question

I know that I have to put it in the x-k form
F(x)=(x-1)(x-3)(x+2)
and then multiply them but I dont know the y-intercept! Please help me
Thanks

Answer 1

You're almost there. With the polynomial you propose, you've covered (1,0)(3,0)(-2,0)
But now, let me propose this polynomial:
f(x) = a(x-1)(x-3)(x+2)
We have that f(2) = 1, so:
f(2) = a(2-1)(2-3)(2+2) = -4a = 1. Then a = -1/4

The function you need is

f(x) = -1/4(x-1)(x-3)(x+2)

Answer 2

F(x)=C(x-1)(x-3)(x+2)
Plug in (2,1),
1 = -4C
C= -1/4
Therefore, the cubic function is
F(x)=-(1/4)(x-1)(x-3)(x+2)

Answer 3

for (1,0)(3,0)(-2,0) the Y intercept is Zero, but for (2,1) it is 1. i hope you can solve this problem. i haven't done these kind of problems" I think". but i can look it up for you.
bye and good luck!

Answer 4

y = -0.25x^3 + 0.5x^2 + 1.25x - 1.5
works out.

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