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 $x=-3$.