What would this function be?

Question

Your function should have roots -2, -1, and 2 also must include the point (0,4).

Answer 1

Here is a similar question: What function has roots of 3, 6, and -9 and includes the point (0,81)?

If the function is to be polynomial (and this is the importance of context, eh?), then it must have factors corresponding to its roots, in this case: (x - 3), (x - 6), and (x + 9). It can also have a constant factor, k. Thus our function can be modelled as: f(x) = k(x - 3)(x - 6)(x + 9). In order for (0,81) to be a solution, we replace x with 0 and also substitute 81 for f(x): 81 = k(0 - 3)(0 - 6)(0 + 9). Now we solve for k.

81 = k (18)(9)

k = 1/2

Therefore our function is f(x) = (1/2)(x - 3)(x - 6)(x + 9).

That is the simplest function that will do. There are an infinite variety of other polynomial functions that will also work because any of the binomial factors can be raised to a natural power without changing the requirements that this fulfills -- though that will require us to recalculate the constant factor. For example, g(x) = c(x - 3)^2 (x - 6) (x + 9) also has roots of 3, 6, and -9. If its graph is to have a point at (0,81), then

81 = c (-3)^2 (-6) (9)

81 = c (-486)

c = -1/6

So g(x) = (-1/6) (x - 3)^2 (x - 6) (x + 9) will also do!

Answer 2

Using the zeros provided, the factors equation would use the following factors:

y=a(x+2)(x+1)(x-2)

To determine what I is, you would plug in the point provided (0,4)

4=a(0+2)(0+1)(0-2)

4=a(2)(1)(-2)\

4=-4a Divide both sides by -4

4/-4=-4a/-4

-1=a

Plugging this in

y= -(x+2)(x+1)(x-2)