Polynomials Question 1

Which of the following could be an equation for the graph shown in the $xy$ -plane above?

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

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

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

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

Approach 1

Using the graph, we identify the following zeroes (x-intercepts):

A polynomial with zeroes at $x=-2,0,3$ must have the following factors. Because the graph bounces off the point $(0,0)$ , the exponent of $x$ must be even.

y=(x)^{\text{any even power}}(x+2)^{\text{any odd power}}(x-3)^{\text{any odd power}}

Only one answer choice meets these requirements.

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

\boxed{y=x^2(x+2)(x-3)}

Approach 2

We may choose an arbitrary point to test. For example, if we plug in $x=2$ , we expect a negative value. The first and third answer choices would be incompatible since they evaluate to 0.

Choosing another arbitrary point, such as $x=-3$ , we expect a positive value. Only the last option is compatible when we plug in $x=-3$ .