To find the x-intercepts of the function \( f(x) = (x + 9)^2 (3x + 1) \), we need to set the function equal to zero and solve for \( x \):
\[ f(x) = (x + 9)^2 (3x + 1) = 0 \]
The product of the factors is zero if either factor is zero. We need to find the values of \( x \) for which each factor is zero:
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\( (x + 9)^2 = 0 \)
- This gives us \( x + 9 = 0 \) or \( x = -9 \).
- Since this is a squared term, \( x = -9 \) is a repeated root, meaning it counts as one x-intercept.
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\( (3x + 1) = 0 \)
- This gives us \( 3x + 1 = 0 \) or \( 3x = -1 \), so \( x = -\frac{1}{3} \).
- This is a simple root, so it counts as one x-intercept.
Now we tally the x-intercepts:
- From \( (x + 9)^2 \) we have one x-intercept at \( x = -9 \).
- From \( (3x + 1) \) we have one x-intercept at \( x = -\frac{1}{3} \).
In total, the function has two x-intercepts: one at \( x = -9 \) (with multiplicity 2) and one at \( x = -\frac{1}{3} \).
Thus, the answer is that the function has 2 x-intercepts.