To determine the intervals where the function [tex]\( f(x) = x^2 - 3x - 4 \)[/tex] is negative, we need to follow these steps:
1. Find the roots of the quadratic equation: To identify where the function intersects the x-axis, we solve [tex]\( x^2 - 3x - 4 = 0 \)[/tex].
2. Solve for the roots: The quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has roots given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}
\][/tex]
This gives us:
[tex]\[
x = \frac{3 + 5}{2} = 4 \quad \text{and} \quad x = \frac{3 - 5}{2} = -1
\][/tex]
3. Determine the sign of the function in each interval: The roots we found divide the real number line into three intervals:
- [tex]\( (-\infty, -1) \)[/tex]
- [tex]\( (-1, 4) \)[/tex]
- [tex]\( (4, \infty) \)[/tex]
We need to test the sign of [tex]\( f(x) \)[/tex] in each of these intervals.
- For [tex]\( (-\infty, -1) \)[/tex]: Choose a test point such as [tex]\( x = -2 \)[/tex].
[tex]\[
f(-2) = (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 \quad (\text{positive})
\][/tex]
- For [tex]\( (-1, 4) \)[/tex]: Choose a test point such as [tex]\( x = 0 \)[/tex].
[tex]\[
f(0) = 0^2 - 3 \cdot 0 - 4 = -4 \quad (\text{negative})
\][/tex]
- For [tex]\( (4, \infty) \)[/tex]: Choose a test point such as [tex]\( x = 5 \)[/tex].
[tex]\[
f(5) = 5^2 - 3 \cdot 5 - 4 = 25 - 15 - 4 = 6 \quad (\text{positive})
\][/tex]
Based on the test points above, [tex]\( f(x) \)[/tex] is negative only in the interval [tex]\( (-1, 4) \)[/tex].
Thus, the correct answer is:
II only
