Question

which equation choice could represent the graph shown below?
answer
$f(x) = (x + 8)(x^2 - 9)$
$f(x) = (x - 8)(x^2 - 9)$
$f(x) = (x - 8)(x^2 + 9)$
$f(x) = (x + 8)(x^2 + 9)$

Explanation:

Step1: Find x - intercept

The graph intersects the x - axis at \(x=-8\) (since it crosses the x - axis at \(x = - 8\)). For a function \(f(x)\) in factored form, if \(x = a\) is an x - intercept, then \((x - a)\) is a factor. So when \(x=-8\), \(x + 8=0\), so \((x + 8)\) should be a factor. This eliminates the options \(f(x)=(x - 8)(x^{2}-9)\) and \(f(x)=(x - 8)(x^{2}+9)\) because they have \((x - 8)\) as a factor (which would correspond to an x - intercept at \(x = 8\), but our graph has an x - intercept at \(x=-8\)).

Step2: Analyze the quadratic factor

Now we have two remaining options: \(f(x)=(x + 8)(x^{2}-9)\) and \(f(x)=(x + 8)(x^{2}+9)\). Let's analyze the quadratic part. The quadratic \(x^{2}+9\) has discriminant \(\Delta=0^{2}-4\times1\times9=- 36<0\), so it has no real roots. The quadratic \(x^{2}-9=(x - 3)(x + 3)\) has real roots \(x = 3\) and \(x=-3\). But our graph only has one x - intercept (at \(x=-8\)), which means the other factor should not have real roots? Wait, no, wait. Wait, the graph shown has only one x - intercept? Wait, no, looking at the graph, it crosses the x - axis at \(x=-8\) and maybe the other factor? Wait, no, the graph as shown: let's check the end behavior. The leading term of \(f(x)=(x + 8)(x^{2}-9)=x^{3}+8x^{2}-9x - 72\), 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. For \(f(x)=(x + 8)(x^{2}+9)=x^{3}+8x^{2}+9x + 72\), leading term \(x^{3}\), but the quadratic \(x^{2}+9\) has no real roots, so the only x - intercept would be at \(x=-8\). But wait, the graph in the picture: let's see the y - intercept. Let's compute the y - intercept for \(f(x)=(x + 8)(x^{2}-9)\): when \(x = 0\), \(f(0)=(0 + 8)(0 - 9)=8\times(-9)=-72\)? Wait, no, the graph in the picture crosses the y - axis at positive y? Wait, maybe I made a mistake. Wait, no, let's re - check. Wait, \(x^{2}-9=(x - 3)(x + 3)\), so \(f(x)=(x + 8)(x - 3)(x + 3)\). The roots are \(x=-8\), \(x = 3\), \(x=-3\). But the graph shown has only one x - intercept? Wait, maybe the graph is drawn in a way that the other intercepts are not visible? Wait, no, the x - axis is from - 10 to 10. The graph crosses the x - axis at \(x=-8\) and maybe the other roots are outside? Wait, no, \(x = 3\) and \(x=-3\) are within - 10 to 10. Wait, maybe the graph is a cubic with one real root? Wait, no, a cubic must have at least one real root. Wait, the function \(f(x)=(x + 8)(x^{2}+9)\) has one real root at \(x=-8\) and two complex roots. The function \(f(x)=(x + 8)(x^{2}-9)\) has three real roots. But the graph shown has only one x - intercept (at \(x=-8\)). So that would mean the other factor has no real roots. So \(x^{2}+9\) has no real roots, so the function \(f(x)=(x + 8)(x^{2}+9)\) has only one real root at \(x=-8\). But wait, the end behavior: for \(f(x)=(x + 8)(x^{2}+9)=x^{3}+8x^{2}+9x + 72\), as \(x ightarrow\infty\), \(f(x) ightarrow\infty\) and as \(x ightarrow-\infty\), \(f(x) ightarrow-\infty\), which matches the graph. For \(f(x)=(x + 8)(x^{2}-9)\), as \(x ightarrow\infty\), \(f(x) ightarrow\infty\) and as \(x ightarrow-\infty\), \(f(x) ightarrow-\infty\) (same end behavior). But the graph has only one x - intercept, so the correct function should be the one with only one real root. So \(f(x)=(x + 8)(x^{2}+9)\)? Wait, no, wait, maybe I misread the graph. Wait, the original graph: let's check the y - intercept. For \(f(x)=(x + 8)(x^{2}+9)\), when \(x = 0\), \(f(0)=(0 + 8)(0 + 9)=72\), which is positive. For \(f(x)=(x + 8)(x^{2}-9)\), when \(x = 0\), \(f(0)=(0 + 8)(0 - 9)=-72\), which is negative. The graph shown crosses the y - axis at positive y, so \(f(0)>0\). So \(f(x)=(x + 8)(x^{2}+9)\) has \(f(0)=72>0\), and \(f(x)=(x + 8)(x^{2}-9)\) has \(f(0)=-72<0\). So the correct function should be \(f(x)=(x + 8)(x^{2}+9)\)? Wait, but earlier I thought the other factor. Wait, no, maybe the graph has only one x - intercept, so the quadratic factor has no real roots. So the correct option is \(f(x)=(x + 8)(x^{2}+9)\)? Wait, no, wait the options:

Wait, let's re - evaluate. The graph intersects the x - axis at \(x=-8\). So when \(x=-8\), \(f(x)=0\). Let's check each option:

1. \(f(x)=(x + 8)(x^{2}-9)\): \(f(-8)=0\), good. \(f(8)=(16)(64 - 9)=16\times55 eq0\), so x - intercept at \(x=-8\), \(x = 3\), \(x=-3\). But the graph shows only one x - intercept? Maybe the graph is drawn with only \(x=-8\) as the visible x - intercept, but the other roots are outside the graph's window? Wait, the x - axis is from - 10 to 10. \(x = 3\) and \(x=-3\) are within - 10 to 10. But the graph as shown doesn't cross the x - axis at \(x = 3\) or \(x=-3\). So that suggests that the other factor has no real roots. So the quadratic factor should be \(x^{2}+9\) (which has no real roots). So the function is \(f(x)=(x + 8)(x^{2}+9)\). Wait, but let's check the y - intercept again. \(f(0)=(0 + 8)(0 + 9)=72>0\), which matches the graph (since it crosses the y - axis at positive y). The function \(f(x)=(x + 8)(x^{2}-9)\) has \(f(0)=-72<0\), which doesn't match. So the correct answer is \(f(x)=(x + 8)(x^{2}+9)\).

Wait, but wait, the leading term of \(f(x)=(x + 8)(x^{2}+9)=x^{3}+8x^{2}+9x + 72\), the degree is 3, leading coefficient positive, so as \(x ightarrow\infty\), \(f(x) ightarrow\infty\) and as \(x ightarrow-\infty\), \(f(x) ightarrow-\infty\), which matches the graph's end behavior (going down as \(x ightarrow-\infty\) and up as \(x ightarrow\infty\)). The function \(f(x)=(x + 8)(x^{2}-9)\) also has leading term \(x^{3}\), same end behavior. But the y - intercept: the graph crosses the y - axis at positive y, so \(f(0)>0\). So \(f(x)=(x + 8)(x^{2}+9)\) has \(f(0)=72>0\), which is correct. The other option \(f(x)=(x + 8)(x^{2}-9)\) has \(f(0)=-72<0\), which is incorrect. So the correct answer is \(f(x)=(x + 8)(x^{2}+9)\).

Answer:

\(f(x)=(x + 8)(x^{2}+9)\)