To write a possible equation for a polynomial whose graph has horizontal intercepts at $x = 2$, $x = -\frac{1}{2}$, and $x = -3$, we construct the equation using the roots. The polynomial equation can be expressed as:

$P(x) = a(x - 2)\left(x + \frac{1}{2}\right)(x + 3)$

Here, $a$ is a constant that can scale the polynomial but does not change the locations of the intercepts.

Simplify the equation:

1. Expand $x + \frac{1}{2}$: $x + \frac{1}{2} = \frac{2x + 1}{2}$

   So the polynomial becomes: $P(x) = a(x - 2)\left(\frac{2x + 1}{2}\right)(x + 3)$

2. Simplify by factoring out the denominator: $P(x) = \frac{a}{2}(x - 2)(2x + 1)(x + 3)$

3. Expand the terms if needed. Multiplying $(2x + 1)(x + 3)$: $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$

   Now multiply this result by $x - 2$: $P(x) = \frac{a}{2}(x - 2)(2x^2 + 7x + 3)$

   Expand: $P(x) = \frac{a}{2}(2x^3 + 7x^2 + 3x - 4x^2 - 14x - 6)$

   Combine like terms: $P(x) = \frac{a}{2}(2x^3 + 3x^2 - 11x - 6)$

4. Final simplified polynomial: $P(x) = a(x^3 + \frac{3}{2}x^2 - \frac{11}{2}x - 3)$

A simple example is to let $a = 2$, making the leading coefficient an integer: $P(x) = 2(x^3 + \frac{3}{2}x^2 - \frac{11}{2}x - 3)$

Do you have any questions or want further clarification?

Related Questions:

1. What does the coefficient $a$ represent in the polynomial equation?
2. How do horizontal intercepts relate to the roots of a polynomial?
3. Can there be multiple polynomial equations with the same intercepts? If yes, how are they related?
4. How would you adjust the polynomial if the root $x = 2$ had a multiplicity of 2?
5. What is the role of the degree of a polynomial in determining the number of horizontal intercepts?

Tip:

Always verify the intercepts by substituting them back into the equation to ensure the polynomial equals zero at those points.