Question

select the correct answer.
which equation could possibly represent the graphed function?
a. $f(x)=(x - 4)(x + 2)(x + 4)$
b. $f(x)=(x - 4)^2(x - 2)$
c. $f(x)=(x + 4)^2(x + 2)$
d. $f(x)=(x - 4)(x - 2)(x + 4)$

Explanation:

Step1: Identify x-intercepts from graph

The graph crosses the x-axis at $x=-2$ and $x=4$, and touches (bounces off) the x-axis at $x=-4$.

Step2: Match intercepts to factors

For a root $a$, the corresponding factor is $(x-a)$. A repeated root (touching intercept) has an even exponent. So factors are $(x+4)^2$ (for $x=-4$) and $(x+2)$ (for $x=-2$), and $(x-4)$ (for $x=4$)? No, wait: check end behavior. As $x\to+\infty$, $f(x)\to+\infty$; as $x\to-\infty$, $f(x)\to-\infty$. This means leading coefficient is positive, and degree is odd. Now check options:

• Option A: $f(x)=(x-4)(x+2)(x+4)$: roots at 4, -2, -4, all single. But graph has a bounce at $x=-4$, so this is wrong.
• Option B: $f(x)=(x-4)^2(x-2)$: roots at 4 (double), 2. Does not match graph's intercepts.
• Option C: $f(x)=(x+4)^2(x+2)$: roots at -4 (double), -2. Missing $x=4$ intercept, wrong.
• Option D: $f(x)=(x-4)(x-2)(x+4)$: roots at 4, 2, -4. No, wait correction: graph's intercepts are $x=-4$ (bounce), $x=-2$, $x=4$. Wait, recheck: when $x=4$, the graph goes to $+\infty$ on right, $-\infty$ on left. Let's test $x=0$ on graph: $f(0)$ is between -30 and -40.

Test Option A: $f(0)=(0-4)(0+2)(0+4)=(-4)(2)(4)=-32$, which matches the graph's $f(0)$ value. Wait, but the bounce? No, maybe misread the graph: the left intercept is $x=-4$, crosses, not bounces? Wait no, Option D: $f(0)=(0-4)(0-2)(0+4)=(-4)(-2)(4)=32$, positive, which does not match graph (f(0) is negative). Option A: $f(0)=-32$, negative, matches. Option A's roots: $x=4, -2, -4$. The graph crosses at all three? Maybe the left curve is a steep cross, not a bounce. Let's confirm end behavior: Option A is cubic, leading term $x^3$, so as $x\to+\infty$, $f(x)\to+\infty$; $x\to-\infty$, $f(x)\to-\infty$, which matches the graph.

Step3: Verify intercepts and value

Graph has x-intercepts at $x=-4, x=-2, x=4$, which matches Option A's roots. $f(0)=-32$, which aligns with the graph's y-value at x=0 being negative between -30 and -40.

Answer:

A. $f(x) = (x - 4)(x + 2)(x + 4)$