Question

the graph of the function $f(x) = (x + 2)(x + 6)$ is shown below.
which statement about the function is true?

the function is positive for all real values of $x$ where $x > -4$.
the function is negative for all real values of $x$ where $-6 < x < -2$.
the function is positive for all real values of $x$ where $x < -6$ or $x > -3$.
the function is negative for all real values of $x$ where $x < -2$.

Explanation:

Step1: Analyze the roots of the function

The function \( f(x)=(x + 2)(x + 6) \) is a quadratic function. The roots are found by setting \( f(x)=0 \), so \( x+2 = 0 \) gives \( x=-2 \) and \( x + 6=0 \) gives \( x=-6 \). The parabola opens upwards (since the coefficient of \( x^{2} \) when expanded is positive, \( f(x)=x^{2}+8x + 12 \), coefficient of \( x^{2} \) is \( 1>0 \)).

Step2: Analyze the sign of the function in intervals

• For \( x<-6 \): Let's take \( x=-7 \), then \( f(-7)=(-7 + 2)(-7+6)=(-5)\times(-1) = 5>0 \), so the function is positive.
• For \( -6
• For \( x>-2 \): Let's take \( x = 0 \), then \( f(0)=(0 + 2)(0+6)=12>0 \), so the function is positive.

Now let's check each option:

• Option 1: "The function is positive for all real values of \( x \) where \( x>-4 \)". For \( -4 \) is between \( -6 \) and \( -2 \), when \( -6
• Option 2: "The function is negative for all real values of \( x \) where \( -6
• Option 3: "The function is positive for all real values of \( x \) where \( x<-6 \) or \( x>-3 \)". When \( x>-3 \), if \( -3 \) is between \( -6 \) and \( -2 \) (wait, \( -3 \) is in \( -6-2 \) the function is positive. But the interval \( x>-3 \) includes \( -3
• Option 4: "The function is negative for all real values of \( x \) where \( x<-2 \)". But for \( x<-6 \), the function is positive, so this is false.

Answer:

The function is negative for all real values of \( x \) where \( -6 < x < -2 \) (the second option).