Question

the function $f(x) = -(x + 5)(x + 1)$ is shown.
what is the range of the function?
\bigcirc all real numbers less than or equal to 4
\bigcirc all real numbers less than or equal to -3
\bigcirc all real numbers greater than or equal to 4
\bigcirc all real numbers greater than or equal to -3

Explanation:

Step1: Analyze the parabola's direction

The function is \( f(x) = -(x + 5)(x + 1) \). The coefficient of \( x^2 \) (after expanding) will be negative (since the leading factor is -1), so the parabola opens downward.

Step2: Find the vertex (maximum point)

First, expand the function:
\( f(x) = -(x^2 + 6x + 5) = -x^2 - 6x - 5 \).
The x - coordinate of the vertex of a parabola \( ax^2+bx + c \) is \( x = -\frac{b}{2a} \). Here, \( a=-1 \), \( b = -6 \), so \( x=-\frac{-6}{2\times(-1)}=-3 \).

Now, substitute \( x = -3 \) into \( f(x) \):
\( f(-3)=-(-3 + 5)(-3 + 1)=-(2)(-2)=4 \). So the vertex is at \( (-3, 4) \), which is the maximum point.

Step3: Determine the range

Since the parabola opens downward, the maximum value of \( y \) (output of the function) is 4, and \( y \) can take any real number less than or equal to 4.

Answer:

all real numbers less than or equal to 4