Question

How do I solve this ​

Answer 1

The function f(x)=a(x+3)(x-2)(x-4) has those zeros. Plug in the coordinates and solve for a=2.

f(x)=2(x+3)(x-2)(x-4)
You can also multiply that out if you want to

Answer 2

f(x) = 2x^3 - 6x^2 - 20x + 48

Explanation:

A polynomial with zeros a, b, c, etc., is the product of (x - a)(x - b)(x - c)...

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

f(x) = (x^2 + x - 6)(x - 4)

f(x) = x^3 - 4x^2 + x^2 - 4x - 6x + 24

f(x) = x^3 - 3x^2 - 10x + 24

This polynomial function has the zeros listed in the problem.

Now we need to make sure it includes the point (6, 144).

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

f(6) = (6 + 3)(6 - 2)(6 - 4)

f(6) = 9 * 4 * 2

f(6) = 72

The polynomial function has point (6, 72). We want it to include the point (6, 144). We multiply the function by 2.

f(x) = 2(x^3 - 3x^2 - 10x + 24)

f(x) = 2x^3 - 6x^2 - 20x + 48