How many x-intercepts would the function shown below have?

f, of, x, equals, left parenthesis, x, plus, 9, right parenthesis, squared, left parenthesis, 3, x, plus, 1, right parenthesis
f(x)=(x+9)
2
(3x+1)

1 answer

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:

  1. \( (x + 9)^2 = 0 \)

    • 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.
  2. \( (3x + 1) = 0 \)

    • 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.

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