math problem?

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 

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  • 4 weeks ago

    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

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