Let the radius of sphere Q be $R$. Using the surface area formula, how can you write that the surface area of Q is $144\pi$ greater than that of P?

From your surface area equation, solve for $R^2$ in terms of $x$. What constant gets added to $x^2$?

Once you have $R$ in terms of $x$, plug it into the volume formula $V = \frac{4}{3}\pi R^3$ and simplify the exponent so it matches one of the answer choices.

In one Desmos expression line, type

This expression represents $R^2$. Let Desmos simplify it; the simplified result is the formula for $R^2$ in terms of $x$.

In a new line, define the volume function using that result. For example, if Desmos shows $R^2 =$ some expression $E(x)$, then type:

This is your candidate formula for the volume of sphere Q as a function of $x$.

Now type the right-hand side of each answer choice as separate functions, such as $A(x)$, $B(x)$, $C(x)$, and $D(x)$. Look to see which graph lies exactly on top of $V(x)$ for all positive $x$; that option is the one that matches your derived function.

For a sphere with radius $r$:

• Surface area: $S = 4\pi r^2$
• Volume: $V = \frac{4}{3}\pi r^3$

Sphere P has radius $x$, so its surface area is $4\pi x^2$. Let the radius of sphere Q be $R$; its surface area is then $4\pi R^2$ and its volume is $\frac{4}{3}\pi R^3$.

We are told that sphere Q’s surface area is $144\pi$ greater than sphere P’s surface area.

So:

Divide both sides by $4\pi$:

This expresses $R^2$ in terms of $x$.

Since $x>0$ and $R$ is a radius (so it must be positive), we take the positive square root:

The volume of sphere Q is

Substitute $R$ to get a volume function in terms of $x$ only.

Substitute $R = \sqrt{x^2 + 36}$ into the volume formula:

Now rewrite the cube of a square root:

So the volume of sphere Q as a function of $x$ is

which matches answer choice B.