Question

1. Given f of x is equal to the quantity x minus 7 end quantity divided by the quantity x squared minus 3x minus 28 end quantity, which of the following is true?

f(x) is positive for all x < 7
f(x) is negative for all x < 7
f(x) is positive for all x > –4
f(x) is negative for all x > –4

Answer 1

To analyze the function:

\[
f(x) = \frac{x - 7}{x^2 - 3x - 28}
\]

we first need to factor the denominator. The quadratic expression \(x^2 - 3x - 28\) can be factored as follows:

1. We look for two numbers that multiply to \(-28\) and add to \(-3\). The numbers \(-7\) and \(4\) fit these criteria, so we can factor it as:
\[
x^2 - 3x - 28 = (x - 7)(x + 4)
\]

Thus, we have:
\[
f(x) = \frac{x - 7}{(x - 7)(x + 4)}
\]

Now, we can simplify \(f(x)\), noting that we must be cautious of points where the denominator equals zero:

\[
f(x) = \frac{1}{x + 4}, \quad \text{for } x \neq 7
\]

Next, we check the behavior of \(f(x)\) on the specified intervals.

### 1. For \(x < 7\):

- Let's analyze \(f(x)\) when \(x < 7\):
- If \(x < -4\), then \(x + 4 < 0\), so \(f(x) = \frac{1}{x + 4} < 0\).
- If \(-4 < x < 7\), then \(x + 4 > 0\), so \(f(x) = \frac{1}{x + 4} > 0\).

### 2. For \(x > -4\):

- If \(-4 < x < 7\), we already established that \(f(x) > 0\).
- If \(x > 7\), then \(f(x) = \frac{1}{x + 4} > 0\).

### Summary of results:

- For \(x < -4\), \(f(x)\) is negative.
- For \(-4 < x < 7\), \(f(x)\) is positive.
- For \(x > 7\), \(f(x)\) is positive.

### Conclusion:

The correct choices from the options you provided are:

- \(f(x)\) is negative for all \(x < -4\).
- \(f(x)\) is positive for all \(x > -4\).

The only conclusion we can definitively support from the options is:
**f(x) is negative for all x < -4.**

However, since this specific option is not provided, the next closest evaluation we can assert from the given options is:
- **\(f(x)\) is positive for all \(x > -4\)**.