Questions: Question 21 4 pts Use the given table for (f(x)=a x^2+b x+c) to solve the inequality (f(x)<0). x -6 -4 -2 0 2 4 6 f(x) 20 0 -12 -16 -12 0 20 (x<-4) or (x>4) (x<-5) or (x>5) (-5<x<5) (-4<x<4)

Solution Steps

To solve the inequality \( f(x) < 0 \) using the given table, we need to identify the intervals where the function values are negative. By examining the table, we can see that \( f(x) \) is negative for \( x = -2, 0, 2 \). Therefore, the interval where \( f(x) < 0 \) is between the points where the function changes from positive to negative and back to positive. This corresponds to the interval \(-4 < x < 4\).

From the given table, we have the function values \( f(x) \) at specific points:

• \( f(-6) = 20 \)
• \( f(-4) = 0 \)
• \( f(-2) = -12 \)
• \( f(0) = -16 \)
• \( f(2) = -12 \)
• \( f(4) = 0 \)
• \( f(6) = 20 \)

By analyzing the function values, we find that \( f(x) < 0 \) in the intervals:

• Between \( x = -2 \) and \( x = 0 \)
• Between \( x = 0 \) and \( x = 2 \)

The combined interval where \( f(x) < 0 \) is: \[ (-2, 0) \cup (0, 2) \]

Final Answer