To find the intervals in which the function [tex]\( f(x) = x^2 - 7x + 10 \)[/tex] is positive, we will analyze its behavior step-by-step:
1. Identify the roots of the quadratic function:
The quadratic equation [tex]\( f(x) = x^2 - 7x + 10 \)[/tex] can be factored or solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. The equation [tex]\( x^2 - 7x + 10 \)[/tex] can be factored as:
[tex]\[
(x - 2)(x - 5) = 0
\][/tex]
Therefore, the roots are [tex]\( x = 2 \)[/tex] and [tex]\( x = 5 \)[/tex].
2. Determine the behavior of the function around the roots:
For the quadratic function [tex]\( f(x) = x^2 - 7x + 10 \)[/tex], it will change signs at its roots (2 and 5). To determine in which intervals the function is positive, consider the sign of the function between and outside these roots.
3. Analyze the intervals:
The roots split the number line into three intervals:
- Interval I: [tex]\( (-\infty, 2) \)[/tex]
- Interval II: [tex]\( (2, 5) \)[/tex]
- Interval III: [tex]\( (5, \infty) \)[/tex]
4. Test the intervals:
- Pick a test point in interval I, say [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 1^2 - 7 \cdot 1 + 10 = 1 - 7 + 10 = 4 \quad (\text{Positive})
\][/tex]
- Pick a test point in interval II, say [tex]\( x = 3 \)[/tex]:
[tex]\[
f(3) = 3^2 - 7 \cdot 3 + 10 = 9 - 21 + 10 = -2 \quad (\text{Negative})
\][/tex]
- Pick a test point in interval III, say [tex]\( x = 6 \)[/tex]:
[tex]\[
f(6) = 6^2 - 7 \cdot 6 + 10 = 36 - 42 + 10 = 4 \quad (\text{Positive})
\][/tex]
5. Determine the intervals in which the function is positive:
Based on the tests above, [tex]\( f(x) \)[/tex] is positive in intervals I and III:
[tex]\[
(-\infty, 2) \quad \text{and} \quad (5, \infty)
\][/tex]
Therefore, the function [tex]\( f(x) = x^2 - 7x + 10 \)[/tex] is positive in the intervals [tex]\( (-\infty, 2) \)[/tex] and [tex]\( (5, \infty) \)[/tex]. The correct choices are:
- I. [tex]\( (-\infty, 2) \)[/tex]
- III. [tex]\( (5, \infty) \)[/tex]
Thus, the answer is:
I and III
