Question

a quadratic function f(x) is hidden from view. you must find all intervals where f(x) is negative. choose the form of the quadratic function f(x) that you would like to see in order to answer the question most efficiently. form: factored form f(x) = -2(x - 3)(x - 1) answer attempt 1 out of 2

Explanation:

Step1: Find roots of the function

To find where \( f(x) = -2(x - 3)(x - 1) \) is negative, first find the roots by setting \( f(x) = 0 \).
\( -2(x - 3)(x - 1)=0 \) implies \( x - 3 = 0 \) or \( x - 1 = 0 \), so \( x = 3 \) or \( x = 1 \).

Step2: Analyze the sign of the function

The coefficient of the quadratic term (when expanded) is \( -2 \), which is negative, so the parabola opens downward.
We can test intervals divided by the roots \( x = 1 \) and \( x = 3 \):

• For \( x < 1 \), let's pick \( x = 0 \). Then \( f(0)=-2(0 - 3)(0 - 1)=-2(-3)(-1)=-6<0 \)? Wait, no, \( -2\times(-3)\times(-1)= - 6\), which is negative? Wait, no, wait: \( (0 - 3)=-3 \), \( (0 - 1)=-1 \), so \( -2\times(-3)\times(-1)=-2\times3=-6 \), which is negative. But wait, the parabola opens downward, so between the roots (1 and 3), the function should be positive, and outside ( \( x < 1 \) or \( x > 3 \) ) it should be negative? Wait, no, let's re - check.

Wait, the factored form is \( f(x)=a(x - r_1)(x - r_2) \), here \( a=-2<0 \), \( r_1 = 1 \), \( r_2 = 3 \).
When \( x < 1 \), say \( x = 0 \): \( (0 - 1)=-1 \), \( (0 - 3)=-3 \), so \( (x - 1)(x - 3)=(-1)\times(-3)=3 \), then \( f(x)=-2\times3=-6<0 \).
When \( 1 < x < 3 \), say \( x = 2 \): \( (2 - 1)=1 \), \( (2 - 3)=-1 \), so \( (x - 1)(x - 3)=1\times(-1)=-1 \), then \( f(x)=-2\times(-1) = 2>0 \).
When \( x>3 \), say \( x = 4 \): \( (4 - 1)=3 \), \( (4 - 3)=1 \), so \( (x - 1)(x - 3)=3\times1 = 3 \), then \( f(x)=-2\times3=-6<0 \).

Wait, but that's a mistake in the initial analysis. Wait, the correct way: for a quadratic function \( y = a(x - r_1)(x - r_2) \) with \( a<0 \), the function is positive between the roots \( r_1 \) and \( r_2 \) and negative outside.
So the roots are \( x = 1 \) and \( x = 3 \), \( a=-2<0 \). So the function is negative when \( x < 1 \) or \( x>3 \), and positive when \( 1 < x < 3 \). Wait, but when we tested \( x = 0 \) (which is \( x < 1 \)), \( f(0)=-6<0 \), when we test \( x = 4 \) ( \( x>3 \) ), \( f(4)=-2(4 - 3)(4 - 1)=-2\times1\times3=-6<0 \), and when we test \( x = 2 \) ( \( 1 < x < 3 \) ), \( f(2)=-2(2 - 3)(2 - 1)=-2\times(-1)\times1 = 2>0 \). So the intervals where \( f(x) \) is negative are \( (-\infty,1)\cup(3,\infty) \).

Wait, I made a mistake earlier in the first test. Let's do it properly:
The critical points are \( x = 1 \) and \( x = 3 \). We divide the real line into three intervals:

1. Interval 1: \( (-\infty,1) \)

• Take \( x = 0 \): \( f(0)=-2(0 - 3)(0 - 1)=-2\times(-3)\times(-1)=-6<0 \)

1. Interval 2: \( (1,3) \)

• Take \( x = 2 \): \( f(2)=-2(2 - 3)(2 - 1)=-2\times(-1)\times1 = 2>0 \)

1. Interval 3: \( (3,\infty) \)

• Take \( x = 4 \): \( f(4)=-2(4 - 3)(4 - 1)=-2\times1\times3=-6<0 \)

Answer:

\((-\infty, 1)\cup(3, \infty)\)