Decreasing Interval Of Transformed Absolute Value Function

Hey guys! Let's dive into a cool math problem involving transformations of the absolute value function. We're given the basic function $f(x)=|x|$, and it's been transformed 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 in algebra and pre-calculus, and once you get the hang of transformations, it's a piece of cake!

First off, let's talk about the original function, $f(x)=|x|$. Remember the graph of $f(x)=|x|$? It looks like a V-shape, with the vertex right at the origin $(0,0)$. For $x gtr 0$, the function is $f(x)=x$, which is a straight line with a positive slope. This means the function is increasing on the interval $(0, gtr)$. For $x < 0$, the function is $f(x)=-x$, which is a straight line with a negative slope. This means the function is decreasing on the interval $(- ginfty, 0)$. So, the key takeaway here is that the absolute value function $f(x)=|x|$ has a minimum at $x=0$, and it decreases to the left of the vertex and increases to the right.

Now, let's break down the transformation from $f(x)=|x|$ to $g(x)=|x+1|-7$. Transformations are like giving our function a makeover! We have two main changes happening here: the '$+1$' inside the absolute value and the '$-7$' outside. The term inside the absolute value, $|x+1|$, affects the horizontal position of the graph. 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. The vertex of $f(x)=|x|$ was at $(0,0)$. After this horizontal shift, the new vertex will be at $(-1,0)$. The other part of the transformation is the '$-7$' outside the absolute value. This is a vertical shift. A '-d' outside the absolute value shifts the graph down by 'd' units. So, $|x+1|-7$ shifts the graph down by 7 units. This means the vertex of our transformed function $g(x)$ is now at $(-1, -7)$.

So, we've taken our V-shaped graph, shifted it one unit left and seven units down. The shape of the 'V' itself remains the same – it's still made of two straight lines meeting at the vertex. The vertex is now at $(-1, -7)$. Because the original function $f(x)=|x|$ decreased for $x<0$ and increased for $x>0$, our transformed function $g(x)=|x+1|-7$ will decrease to the left of its new vertex and increase to the right of its new vertex. The vertex's x-coordinate is what determines the turning point. In this case, the x-coordinate of the vertex of $g(x)$ is $-1$. Therefore, $g(x)$ will be decreasing on the interval to the left of $x=-1$. This interval is $(- ginfty, -1)$.

Let's think about this more formally. The function $g(x) = |x+1| - 7$ can be written as a piecewise function:

If $x+1 gtr 0$ (which means $x > -1$), then $|x+1| = x+1$, so $g(x) = (x+1) - 7 = x - 6$. This part of the function has a slope of 1, meaning it is increasing.

If $x+1 < 0$ (which means $x < -1$), then $|x+1| = -(x+1)$, so $g(x) = -(x+1) - 7 = -x - 1 - 7 = -x - 8$. This part of the function has a slope of -1, meaning it is decreasing.

At $x+1=0$ (which means $x=-1$), the function reaches its minimum value, $g(-1) = |-1+1| - 7 = |0| - 7 = -7$. This is our vertex.

So, the function $g(x)$ is decreasing when $x < -1$. This corresponds to the interval $(- ginfty, -1)$. Looking at the options provided:

A. $(- ginfty,-7)$ B. $(- ginfty,-1)$ C. $(- ginfty, 1)$ D. $(- ginfty, 7)$

Our answer, $(- ginfty, -1)$, matches option B. Awesome!

It's crucial to remember how horizontal shifts work. A common mistake is thinking that $|x+1|$ shifts the graph to the right because of the '+1'. But remember, we're looking for the value of $x$ that makes the expression inside the absolute value zero. For $|x|$, that's $x=0$. For $|x+1|$, that's $x+1=0$, which means $x=-1$. So, the critical point, the vertex, shifts from $x=0$ to $x=-1$. Since the absolute value function is V-shaped and opens upwards, it decreases to the left of the vertex and increases to the right. Therefore, the interval of decrease is always to the left of the x-coordinate of the vertex.

Let's consider another example just to solidify this. If we had $h(x) = |x-3| + 2$, the vertex would be at $(3, 2)$. The expression inside becomes zero when $x-3=0$, so $x=3$. The function $h(x)$ would be decreasing on $(- ginfty, 3)$ and increasing on $(3, ginfty)$. The vertical shift ($+2$ in this case) affects the y-coordinate of the vertex but doesn't change the x-value where the function transitions from decreasing to increasing.

In our original problem, $g(x) = |x+1| - 7$, the vertex is at $x=-1$. The function decreases for all x-values less than $-1$. This is the interval $(- ginfty, -1)$. It's like tracing the graph from left to right. As you move towards the vertex from the left, the y-values are going down. Once you pass the vertex and move to the right, the y-values start going up. So, the decreasing part is definitely on the left side of the vertex.

Understanding transformations is key to mastering these kinds of problems. Remember:

• $|x-h|$ shifts the graph horizontally by $h$ units. If $h$ is positive, it shifts right. If $h$ is negative (like in $|x+1|$, where $h=-1$), it shifts left.
• $k|x|$ stretches or compresses the graph vertically. If $k>1$, it stretches. If $0<k<1$, it compresses. If $k<0$, it also reflects the graph across the x-axis.
• $|x|+k$ shifts the graph vertically by $k$ units. If $k$ is positive, it shifts up. If $k$ is negative, it shifts down.

For $g(x) = |x+1| - 7$, we have a horizontal shift of $-1$ (left by 1) and a vertical shift of $-7$ (down by 7). The vertex moves from $(0,0)$ to $(-1,-7)$. The fundamental V-shape, decreasing to the left of the vertex and increasing to the right, remains the same. Therefore, the function $g(x)$ is decreasing on the interval $(- ginfty, -1)$.

So, to recap, the interval where the function $g(x)=|x+1|-7$ is decreasing is determined by the x-coordinate of its vertex. The vertex occurs where the expression inside the absolute value is zero, i.e., $x+1=0$, which gives $x=-1$. Since the absolute value function opens upwards, it decreases to the left of the vertex. Thus, the function is decreasing on the interval $(- ginfty, -1)$. This corresponds to option B.

Keep practicing, guys, and these transformations will become second nature! Let me know if you have any other tricky math questions!