math problem?

Question

What is a possible equation for the following:a degree three polynomial that has two x-intercepts one at -2 and the other at 3. additionally state the y-int of the equation 

Answer 1

I'm arbitrarily going to pick the first x-intercept to be a double-root, so we end up with 3 roots:

The three x-intercepts are:

x = -2

x = -2

x = 3

Rewrite those as expressions equal to zero:

x + 2 = 0

x + 2 = 0

x - 3 = 0

Multiply those binomials together:

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

That's one possible function, though we could multiply it by any non-zero constant and not affect the x-intercepts:

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

Depending on what we pick for a, it will change the y-intercept. So let's take the simplest case where a = 1 and we just have:

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

Expand that out:

y = (x² + 4x + 4)(x - 3)

y = x(x² + 4x + 4) - 3(x² + 4x + 4)

y = x^3 + 4x² + 4x - 3x² - 12x - 12

y = x^3 + x² - 8x - 12

That's one possible function. The y-intercept is found by setting x=0 which will cancel the first three x terms. So the y-intercept is the constant term (-12).

Answer:

y = x^3 + x² - 8x - 12

y-intercept = -12

P.S. The domain and range of a third degree polynomial is all real numbers. Just ask next time. :)

An alternate answer, just for variety:

y = 2x^3 + 2x² - 16x - 24

y-intercept = -24