Math help please

Question

Third-degree, with zeros of −3, −2, and 1, and passes through the point (4,9).

Answer 1

Third-degree, with zeros of −3, −2, and 1, and passes through the point (4,9).

The zeros tell us that x=-3. x=-2. x=1

You move the numbers to the same side as the x to make them factors and write the equation below

y=a(x+3)(x+2)(x-1)

You take the point you are given (4,9) and substitute the values into the equation to solve for a

9=a(4+3)(4+2)(4-1)

9=a(7)(6)(3)

9=126a

Divide both sides by 126

9/126=a

Reduce

a=1/14

Substitute back in the equation

y=(1/14)(x+3)(x+2)(x-1)