Find The Decreasing Interval Of |x+1|-7

Hey guys, let's dive into a cool transformation problem with absolute value functions! We've got the basic graph of $f(x)=|x|$, and it's been tweaked into a new function, $g(x)=|x+1|-7$. Our mission, should we choose to accept it, is to figure out on which interval this new function, $g(x)$, is decreasing. This is a super common type of question you'll see in algebra and pre-calculus, and understanding transformations is key to crushing it.

First off, let's remind ourselves what the graph of $f(x)=|x|$ looks like. It's a V-shape, with its vertex right at the origin (0,0). For $x e 0$, the graph is made up of two straight lines: $y=x$ for $x>0$ and $y=-x$ for $x<0$. The absolute value function essentially makes any negative input positive. Now, how does this V-shape behave? When $x$ is positive and increasing, $f(x)$ is also positive and increasing. When $x$ is negative and decreasing (moving further left on the number line), $f(x)$ is positive and increasing. So, the function $f(x)=|x|$ is decreasing for $x<0$ and increasing for $x>0$. The vertex at (0,0) is the turning point where the behavior switches. Understanding this basic shape and its increasing/decreasing intervals is the foundation for solving our problem about $g(x)$.

Now, let's talk about the transformations applied to get $g(x)=|x+1|-7$ from $f(x)=|x|$. Transformations are like applying a series of moves to the original graph. In $g(x)$, we have two changes: "+1" inside the absolute value and "-7" outside. The "+1" inside the absolute value, $|x+1|$, causes a horizontal shift. Remember, horizontal shifts work a bit backward to what you might expect. A "+c" inside the absolute value shifts the graph to the left by $c$ units. So, $|x+1|$ shifts the graph of $|x|$ one unit to the left. This means the vertex of our new function will no longer be at (0,0). Instead, the vertex moves to (-1, 0). The "-7" outside the absolute value, $|x+1|-7$, causes a vertical shift. A "-d" outside the function shifts the graph down by $d$ units. So, $|x+1|-7$ shifts the graph down by 7 units. This means the vertex of $g(x)$ is now at (-1, -7).

So, we've taken the V-shape of $f(x)=|x|$, shifted it one unit left, and seven units down to get $g(x)=|x+1|-7$. The V-shape itself hasn't changed, meaning its slope characteristics (except at the vertex) remain the same. The absolute value function, regardless of shifts, always has a vertex and behaves like two linear functions. The key is that the turning point (the vertex) dictates where the function switches from decreasing to increasing, or vice versa. For $f(x)=|x|$, the vertex was at $x=0$. For $g(x)=|x+1|-7$, the vertex occurs when the expression inside the absolute value is zero. That is, when $x+1=0$, which means $x=-1$. So, the vertex of $g(x)$ is at $x=-1$.

Now, let's think about the decreasing interval for $g(x)$. The absolute value function, in its standard form and with vertical shifts, decreases to the left of its vertex and increases to the right of its vertex. Since the vertex of $g(x)=|x+1|-7$ is at $x=-1$, the function will be decreasing for all values of $x$ that are less than -1. This is because as $x$ gets smaller (moves further left on the number line), the value of $|x+1|$ decreases until it reaches its minimum at $x=-1$. The subtraction of 7 just shifts the entire graph down but doesn't change where the turning point is located horizontally. Therefore, the interval on which $g(x)$ is decreasing is $(- ext{infinity}, -1)$. This matches option C. We've successfully analyzed the transformation and determined the decreasing interval!

Understanding the Math Behind the Transformation

Let's really break down why the interval changes the way it does when we transform the absolute value function. The function $g(x) = |x+1| - 7$ is essentially a piecewise function. The definition of absolute value is $|a| = a$ if $a less 0$, and $|a| = -a$ if $a < 0$. Applying this to $g(x)$, we look at the expression inside the absolute value, which is $x+1$.

So, we have two cases:

1. Case 1: $x+1 less 0$ This happens when $x < -1$. In this case, $|x+1| = -(x+1)$. So, $g(x) = -(x+1) - 7 = -x - 1 - 7 = -x - 8$. This is a linear function with a slope of -1. A function with a negative slope is decreasing. So, for all $x < -1$, $g(x)$ is decreasing.

2. Case 2: $x+1 less 0$ This happens when $x less -1$. In this case, $|x+1| = x+1$. So, $g(x) = (x+1) - 7 = x + 1 - 7 = x - 6$. This is a linear function with a slope of +1. A function with a positive slope is increasing. So, for all $x > -1$, $g(x)$ is increasing.

We can see clearly from this piecewise definition that $g(x)$ is decreasing on the interval $(- ext{infinity}, -1)$ and increasing on the interval $(-1, ext{infinity})$. The point $x=-1$ is where the function switches from decreasing to increasing, and it corresponds to the vertex of the V-shaped graph.

The Role of the Vertex

The vertex of the absolute value function is the critical point where the function changes direction. For the basic function $f(x) = |x|$, the vertex is at $(0,0)$. The function decreases for $x<0$ and increases for $x>0$.

When we transform $f(x)$ to $g(x) = |x+1| - 7$, we are performing a horizontal shift and a vertical shift. The horizontal shift is determined by the term inside the absolute value, $|x+1|$. A term of the form $|x-h|$ shifts the graph $h$ units to the right. Conversely, $|x+h|$ (which can be written as $|x-(-h)|$) shifts the graph $h$ units to the left. In our case, $x+1$ means $h=-1$, so the graph shifts 1 unit to the left. This moves the vertex from $x=0$ to $x=-1$.

The vertical shift is determined by the term outside the absolute value, $-7$. A term of the form $|x| - k$ shifts the graph $k$ units down. In our case, $-7$ shifts the graph 7 units down. This moves the vertex from $(0,0)$ to $(-1, -7)$.

Crucially, the vertical shift does not affect the x-coordinate of the vertex, and therefore it does not affect the interval over which the function is decreasing or increasing. The vertex's x-coordinate is solely determined by the horizontal shift. Since the vertex of $g(x)$ is at $x=-1$, the function will be decreasing for all x-values to the left of the vertex, which is the interval $(- ext{infinity}, -1)$.

Comparing with the Options

Let's look at the options provided:

A. $(- ext{infinity}, 7)$ B. $(- ext{infinity}, -7)$ C. $(- ext{infinity}, -1)$ D. $(- ext{infinity}, 1)$

Our analysis clearly shows that the function $g(x) = |x+1| - 7$ is decreasing on the interval $(- ext{infinity}, -1)$. This matches Option C. The other options are incorrect because they either refer to the y-coordinate of the vertex (7 or -7) or the x-coordinate of a different transformation (like $|x-1|$ which would have a vertex at $x=1$). Understanding the effect of horizontal shifts on the vertex's x-coordinate is key to correctly identifying the decreasing interval.

So, to wrap it up, guys, the transformation shifts the V-shape left by 1 unit and down by 7 units. The V-shape itself has a slope of -1 on its left side and +1 on its right side. The turning point, the vertex, is now at $x=-1$. Because the left side has a slope of -1, the function is decreasing on the interval to the left of the vertex, which is $(- ext{infinity}, -1)$. Pretty straightforward once you break down the transformations!