Before dealing with f(2 − x), factor f(x) = x^3 − 4x. Once it is factored, you can easily read off the values of x that make f(x) = 0.

The equation f(2 − x) = 0 means that the input to f is 2 − x. Take each zero of f (each value that makes f equal 0) and set 2 − x equal to that value, then solve for x. Those x-values are the x-intercepts of g(x).

After you find all three distinct x-intercepts of g(x), remember that the question asks for p + q + r, not the individual values. Add the three x-values you found.

In Desmos, enter f(x) = x^3 - 4x to see the base cubic and where it crosses the x-axis. This helps you identify the original zeros of f(x).

Enter g(x) = f(2 - x). Since multiplying by a nonzero constant k only stretches the graph vertically, it does not change the x-intercepts, so this version has the same intercepts as the full g(x).

On the graph of g(x) = f(2 - x), click on each point where the curve crosses the x-axis to read off the three x-coordinates. Then, add those three x-values together to get p + q + r.

The x-intercepts of the graph of $y=g(x)$ occur where $g(x)=0$.

Given

with $k\neq 0$, the equation $g(x)=0$ is equivalent to

Because $k$ is nonzero, this happens exactly when

So to find the x-intercepts of $g$, we just need to solve $f(2-x)=0$. The value of $k$ (and the fact that the leading coefficient is 12) only stretches the graph vertically and does not change the x-intercepts.

First, factor $f(x)$:

From this factorization, $f(x)=0$ when any factor is zero:

• $x=0$
• $x=2$
• $x=-2$

So the zeros of $f$ are $-2$, $0$, and $2$.

Let $t = 2 - x$. Then $f(2-x)=0$ is the same as $f(t)=0$, so $t$ must be one of the zeros of $f$:

Now solve $2-x = t$ for each of these values of $t$:

• If $t=-2$:

• If $t=0$:

• If $t=2$:

Thus, $f(2-x)=0$ when $x=0$, $x=2$, or $x=4$. These are the x-intercepts of $g(x)$, so $p$, $q$, and $r$ are $0$, $2$, and $4$ (in some order).

We now add the three distinct x-intercepts:

So the value of $p+q+r$ is $6$. (Choice D)