Question

which equation choice could represent the graph shown below?
answer
$f(x) = (x + 8)(x^2 + 4)$
$f(x) = (x - 8)(x^2 + 4)$
$f(x) = (x - 8)(x^2 - 4)$
$f(x) = (x + 8)(x^2 - 4)$

Explanation:

Step1: Find x - intercepts

The graph intersects the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, looking at the graph, the x - intercepts are at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, let's check the roots. If a function \(y = f(x)\) has a root at \(x = a\), then \(f(a)=0\). From the graph, the x - intercepts are \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, looking at the graph, the x - intercepts are at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, let's solve for the roots of each option.

For a function \(f(x)=(x - r_1)(x - r_2)(x - r_3)\), the roots are \(x=r_1\), \(x=r_2\), \(x=r_3\).

Let's analyze each option:

Option 1: \(f(x)=(x + 8)(x^{2}+4)\). The roots are found by setting \(f(x)=0\). So \((x + 8)(x^{2}+4)=0\). \(x+8 = 0\) gives \(x=-8\), and \(x^{2}+4=0\) has no real roots (since \(x^{2}=-4\) has no real solutions). But the graph has real roots at \(x=-2\), \(x = 2\), \(x = 8\)? Wait, no, maybe I misread the graph. Wait, the graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, looking at the graph, the x - intercepts are at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the graph: let's see, the graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the grid: the x - axis marks are at - 10, - 8, - 6, - 4, - 2, 0, 2, 4, 6, 8, 10. The graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the rightmost x - intercept is at \(x = 8\), and the left ones at \(x=-2\) and \(x = 2\)? Wait, no, \(x^{2}-4=(x - 2)(x + 2)\), so if we have a factor of \(x^{2}-4\), then the roots are \(x = 2\) and \(x=-2\). And if we have a factor of \((x - 8)\), then the root is \(x = 8\), or \((x + 8)\) with root \(x=-8\).

Wait, let's re - examine the graph. The graph crosses the x - axis at \(x=-2\), \(x = 2\), and \(x = 8\)? Wait, no, the graph: the left x - intercept is at \(x=-2\), middle at \(x = 2\), right at \(x = 8\)? Wait, no, the right x - intercept is at \(x = 8\)? Wait, the grid: the x - axis has a mark at 8, and the graph crosses there. And also at - 2 and 2. So the roots are \(x=-2\), \(x = 2\), \(x = 8\)? Wait, no, \(x=-2\), \(x = 2\) come from \(x^{2}-4=(x - 2)(x + 2)\), and \(x = 8\) comes from \((x - 8)\). Wait, no, let's check option 3: \(f(x)=(x - 8)(x^{2}-4)=(x - 8)(x - 2)(x + 2)\). So the roots are \(x = 8\), \(x = 2\), \(x=-2\), which matches the x - intercepts of the graph (since the graph crosses the x - axis at \(x=-2\), \(x = 2\), \(x = 8\)).

Wait, let's check other options:

Option 2: \(f(x)=(x - 8)(x^{2}+4)\). Roots are \(x = 8\) and \(x^{2}+4 = 0\) (no real roots). But the graph has real roots at \(x=-2\) and \(x = 2\), so this is wrong.

Option 4: \(f(x)=(x + 8)(x^{2}-4)=(x + 8)(x - 2)(x + 2)\). Roots are \(x=-8\), \(x = 2\), \(x=-2\). But the graph has a root at \(x = 8\), not \(x=-8\), so this is wrong.

Option 1: \(f(x)=(x + 8)(x^{2}+4)\). Roots at \(x=-8\) and no real roots from \(x^{2}+4\), which doesn't match the graph.

Option 3: \(f(x)=(x - 8)(x^{2}-4)=(x - 8)(x - 2)(x + 2)\). Roots at \(x = 8\), \(x = 2\), \(x=-2\), which matches the x - intercepts of the graph.

Step2: Verify the end - behavior (optional, but to confirm)

The leading term of \(f(x)=(x - 8)(x^{2}-4)=x^{3}-8x^{2}-4x + 32\). The leading term is \(x^{3}\), so as \(x ightarrow+\infty\), \(f(x) ightarrow+\infty\) and as \(x ightarrow-\infty\), \(f(x) ightarrow-\infty\), which matches the graph (the left end goes down and the right end goes up).

Answer:

\(f(x)=(x - 8)(x^{2}-4)\) (the third option, i.e., the option with \(f(x)=(x - 8)(x^{2}-4)\))