Answer:
zeros are at x = -1, x = 2 and x = -3
continues downward to the left and upward to the right
Step-by-step explanation:
Given function:
f(x) = (x + 1)(x - 2)(x + 3)
The zeros are when f(x) = 0
⇒ (x + 1)(x - 2)(x + 3) = 0
⇒ (x + 1) = 0 ⇒ x = -1
⇒ (x - 2) = 0 ⇒ x = 2
⇒ (x + 3) = 0 ⇒ x = -3
Therefore, the zeros are at x = -1, x = 2 and x = -3
If we expand the brackets of the function we get:
f(x) = (x + 1)(x - 2)(x + 3)
= (x + 1)(x² + x - 6)
= x³ + 2x² - 5x - 6
As the leading coefficient (the coefficient of x³) is positive, the end behavior of the function is:
[tex]f(x) \rightarrow + \infty, \textsf{as } x \rightarrow + \infty[/tex]
[tex]f(x) \rightarrow - \infty, \textsf{as } x \rightarrow - \infty[/tex]
So the correct option is:
"continues downward to the left and upward to the right"
