Question

Review the graph of f(x).

On a coordinate plane, a curve approaches x = negative 3 in quadrant 2 and approaches the x-axis. Another curve approaches x = negative 3 in quadrant 3, has inflection point (0, 0), and approaches x = 3 in quadrant 1. Another cure approaches the x-axis in quadrant 4 and approaches x = 3 in quadrant 4.

Which statement describes the behavior of the function around the vertical asymptotes?

Answer 1

Answer:

Its B

I only answered it on the test and it said it was right

Answer 2

The vertical asymptote concept is used to solve this question. First, I explain the concept, and then, we use it to solve the question.

Using this, the correct option is the second.

Vertical asymptote:  

A vertical asymptote is a value of x for which the function is not defined, that is, it is a point which is outside the domain of a function;  

In a graphic, these vertical asymptotes are given by dashed vertical lines.

In this question:

Vertical asymptoes at $x = -3$ and $x = 3$

At x = -3

To the left(x < -3), goes to plus infinite. To the right(x > -3), goes to negative infitite, so:

$\lim_{x \rightarrow -3^{-}} = \infty, \lim_{x \rightarrow -3^{+}} = -\infty$

At x = 3

To the left(x < 3), goes to plus infinite. To the right(x > 3), goes to negative infinite, so:

$\lim_{x \rightarrow 3^{-}} = \infty, \lim_{x \rightarrow 3^{+}} = -\infty$

Left and right have the same limits both at x = -3 and x = 3, and thus, they behave in the same manner, and the second option is correct.

For more on vertical asymptotes, you can check https://brainly.com/question/23690889