Question

consider the functions.
$f(x) = (x + 1)^2 - 4$
$g(x) = -4|x + 1|$
which statement compares the ranges of the functions?
\bigcirc the ranges of both $f(x)$ and $g(x)$ are limited to $y \geq -1$.
\bigcirc the ranges of both $f(x)$ and $g(x)$ are limited to $y \leq -1$.
\bigcirc since the range of $f(x)$ is limited to $y \leq 0$ and the range of $g(x)$ is limited to $y \geq -4$, the ranges overlap for the values $-4 \leq y \leq 0$.
\bigcirc since the range of $f(x)$ is limited to $y \geq -4$ and the range of $g(x)$ is limited to $y \leq 0$, the ranges overlap for the values $-4 \leq y \leq 0$.

Explanation:

Step1: Analyze the range of \( f(x) = (x + 1)^2 - 4 \)

The function \( f(x) \) is a quadratic function in vertex form \( a(x - h)^2 + k \), where \( a = 1 \), \( h = -1 \), and \( k = -4 \). Since \( a = 1>0 \), the parabola opens upwards, and the vertex is the minimum point. The minimum value of \( (x + 1)^2 \) is \( 0 \) (because the square of any real number is non - negative, i.e., \( (x + 1)^2\geq0 \) for all real \( x \)). So, \( f(x)=(x + 1)^2-4\geq0 - 4=-4 \). Thus, the range of \( f(x) \) is \( y\geq - 4 \).

Step2: Analyze the range of \( g(x)=-4|x + 1| \)

The absolute value function \( |x + 1|\geq0 \) for all real \( x \). When we multiply by \( - 4 \), the inequality sign flips. So, \( -4|x + 1|\leq0 \) (because if \( a\geq0 \), then \( -4a\leq0 \)). Thus, the range of \( g(x) \) is \( y\leq0 \).

Step3: Compare the ranges

The range of \( f(x) \) is \( y\geq - 4 \) and the range of \( g(x) \) is \( y\leq0 \). To find the overlap, we look for the values of \( y \) that satisfy both \( y\geq - 4 \) and \( y\leq0 \). So, the overlapping range is \( -4\leq y\leq0 \).

Answer:

Since the range of \( f(x) \) is limited to \( y\geq - 4 \) and the range of \( g(x) \) is limited to \( y\leq0 \), the ranges overlap for the values \( -4\leq y\leq0 \).