Question

the equation ( v = \frac{4}{3}pi r^3 ) represents the relationship between the volume of a sphere and its radius. what does the end behavior tell you about the relationship between the volume of the sphere and its radius? (1 point)
as the radius decreases to negative infinity, the volume of the sphere will increase to infinity.
as the radius decreases to negative infinity, the volume of the sphere will remain constant.
as the radius increases to infinity, the volume of the sphere will decrease to negative infinity.
as the radius increases to infinity, the volume of the sphere will increase to infinity.

Explanation:

Step1: Analyze the formula

The volume of a sphere is given by \( V=\frac{4}{3}\pi r^{3} \). The radius \( r \) of a sphere is a non - negative quantity (since it represents a length), but we can analyze the function for all real numbers \( r \) in terms of its end - behavior. The function \( V(r)=\frac{4}{3}\pi r^{3} \) is a cubic function with a positive leading coefficient (\( \frac{4}{3}\pi>0 \)).

Step2: Analyze end - behavior as \( r

ightarrow\infty \)
For a cubic function \( y = ax^{3}+bx^{2}+cx + d \) with \( a>0 \), as \( x
ightarrow\infty \), \( y
ightarrow\infty \), and as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \). But in the context of the radius of a sphere, \( r\geq0 \). However, from a mathematical function - analysis perspective, when we consider the end - behavior of the function \( V(r)=\frac{4}{3}\pi r^{3} \):

• When \( r \) increases to infinity (\( r

ightarrow\infty \)):
Since the function is \( V=\frac{4}{3}\pi r^{3} \), and \( \frac{4}{3}\pi>0 \), as \( r \) gets larger and larger (approaches infinity), \( r^{3} \) will also get larger and larger, and multiplying by a positive constant \( \frac{4}{3}\pi \) will make \( V \) approach infinity.

• When \( r \) decreases to negative infinity (\( r

ightarrow-\infty \)):
Since \( r^{3} \) is negative when \( r \) is negative, and \( \frac{4}{3}\pi>0 \), \( V=\frac{4}{3}\pi r^{3} \) will be negative and approach negative infinity as \( r
ightarrow-\infty \). But the radius of a sphere cannot be negative in a physical sense. But among the given options, we analyze the mathematical behavior of the function.
Now let's analyze each option:

• Option 1: "As the radius decreases to negative infinity, the volume of the sphere will increase to infinity."

Since \( V = \frac{4}{3}\pi r^{3} \), if \( r
ightarrow-\infty \), \( r^{3}
ightarrow-\infty \), and \( V=\frac{4}{3}\pi r^{3}
ightarrow-\infty \), so this option is wrong.

• Option 2: "As the radius decreases to negative infinity, the volume of the sphere will remain constant."

Since \( V \) is a function of \( r \), and \( r \) is changing, \( V \) will change. So this option is wrong.

• Option 3: "As the radius increases to infinity, the volume of the sphere will decrease to negative infinity."

Since \( r>0 \) (for a sphere's radius) and \( \frac{4}{3}\pi>0 \), when \( r \) increases, \( r^{3} \) increases, so \( V \) increases. Also, for \( r>0 \), \( V=\frac{4}{3}\pi r^{3}>0 \), so it can't decrease to negative infinity. This option is wrong.

• Option 4: "As the radius increases to infinity, the volume of the sphere will increase to infinity."

Since \( V=\frac{4}{3}\pi r^{3} \), and \( \frac{4}{3}\pi>0 \), when \( r
ightarrow\infty \), \( r^{3}
ightarrow\infty \), so \( V=\frac{4}{3}\pi r^{3}
ightarrow\infty \). This option is correct.

Answer:

As the radius increases to infinity, the volume of the sphere will increase to infinity (the fourth option among the given choices).