SOLUTION: Determine the equation of the polynomial that crosses the x-axis at x=-2 and x=3, bounces off the x-axis at x=1, and has a y-intercept at (0,12)

Question 1110636: Determine the equation of the polynomial that crosses the x-axis at x=-2 and x=3, bounces off the x-axis at x=1, and has a y-intercept at (0,12)
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
This response is not a full solution but only an approach to begin.
Try to sketch your graph's description and you will see the polynomial is degree 4; and since decreasing when unbound to the left and also decreasing when unbound to the right, lead coefficient is negative. You may have something LIKE ;

One of the zeros will be of multiplicity 2.

You may want to start with something in factorized form first...

You might find with the help of a graphing tool that give the right shape and the right zeros, but not yet the right y-intercept, but you can find it by using some "a" as the factor and let x=0 to find "a".
Answer by greenestamps(13216) (Show Source): You can put this solution on YOUR website!

The graph crosses the x-axis (the polynomial has roots) at x=-2 and x=3; that means the polynomial has factors of (x+2) and (x-3).

The graph bounces off the x-axis at x=1; that means the polynomial has a double root at 1, so it has two factors of (x-1).

Note that the multiplicity of the root at x=1 could be any even number; we choose the smallest one for simplicity.

So the equation of the polynomial is

where a is a constant to be determined.

The value of the constant is determined by the y-intercept (0,12); when x=0, y must be 12:

The equation of the polynomial in factored form is

The graph:

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