Question

the function f(x) is graphed below. complete the statements regarding the end behavior.
$f(x) = -4|x + 1| - 3$
show your work here.
hint: to add infinity (∞), type \infinity\
as $x \to -\infty$, $f(x) \to$
as $x \to \infty$, $f(x) \to$

Explanation:

Step1: Analyze the absolute value function

The function is \( f(x) = -4|x + 1| - 3 \). The general form of an absolute value function is \( y = a|x - h| + k \), where \( a \) determines the direction and steepness. Here, \( a = -4 \), which is negative, so the graph opens downward.

Step2: Determine the end behavior as \( x \to -\infty \)

For an absolute value function opening downward (\( a < 0 \)), as \( x \) approaches \( -\infty \), we look at the term \( |x + 1| \). As \( x \to -\infty \), \( |x + 1| \approx |x| \to \infty \). Then \( -4|x + 1| \to -4(\infty) = -\infty \), and subtracting 3 still gives \( -\infty \). Wait, no—wait, \( a = -4 \), so \( -4|x + 1| \) when \( x \to -\infty \): \( |x + 1| \) is positive and large, so \( -4 \times \) large positive is large negative. Then \( -4|x + 1| - 3 \to -\infty \)? Wait, no, wait the graph: the vertex is at \( (-1, -3) \), and it opens downward. Wait, no, let's re-express. Wait, the absolute value \( |x + 1| \) is always non-negative, so \( -4|x + 1| \) is always non-positive (since -4 times non-negative is non-positive), and then subtract 3, so \( f(x) \) is always \( \leq -3 \). Now, as \( x \) goes to \( -\infty \), \( |x + 1| \) increases without bound, so \( -4|x + 1| \) decreases without bound (since it's negative and the magnitude increases), so \( f(x) = -4|x + 1| - 3 \) also decreases without bound? Wait, no, wait: if \( a \) is negative, the graph opens downward, so the ends go downward. Wait, no, absolute value function: when \( a < 0 \), the graph is a "V" opening downward. So as \( x \to \pm\infty \), the function values go to \( -\infty \)? Wait, no, let's take the limit. Let's consider \( x \to \infty \): \( |x + 1| \to \infty \), so \( -4|x + 1| \to -\infty \), so \( f(x) \to -\infty \). Similarly, \( x \to -\infty \): \( |x + 1| = |-(|x| - 1)| = |x| - 1 \) (for large negative \( x \)), so \( |x + 1| \to \infty \), so \( -4|x + 1| \to -\infty \), so \( f(x) \to -\infty \). Wait, but that can't be. Wait, the graph: the vertex is at \( (-1, -3) \), and it opens downward, so both ends (as \( x \to \infty \) and \( x \to -\infty \)) go to \( -\infty \). Wait, but let's check with the function. Let's take \( x = 100 \): \( f(100) = -4|101| - 3 = -404 - 3 = -407 \). \( x = -100 \): \( f(-100) = -4| -99| - 3 = -396 - 3 = -399 \). So as \( x \) becomes very large positive or very large negative, \( f(x) \) becomes very large negative. So both as \( x \to -\infty \) and \( x \to \infty \), \( f(x) \to -\infty \). Wait, but the problem is asking for the end behavior. Wait, maybe I made a mistake. Wait, the absolute value function: the standard \( |x| \) opens upward, \( -|x| \) opens downward. So \( -4|x + 1| - 3 \) is a downward opening V, so as \( x \to \pm\infty \), the function tends to \( -\infty \).

Step3: Confirm the end behavior

So for \( x \to -\infty \), \( f(x) \to -\infty \) (or "negative infinity"), and for \( x \to \infty \), \( f(x) \to -\infty \) (or "negative infinity").

Answer:

As \( x \to -\infty \), \( f(x) \to -\infty \) (or "negative infinity"); as \( x \to \infty \), \( f(x) \to -\infty \) (or "negative infinity").

So filling in the blanks:

For \( x \to -\infty \), \( f(x) \to \boldsymbol{-\infty} \) (or "negative infinity");

For \( x \to \infty \), \( f(x) \to \boldsymbol{-\infty} \) (or "negative infinity").