Graphing Absolute Value Functions And Decreasing Intervals

Hey guys! Today, we're diving into the fascinating world of graphing absolute value functions and pinpointing the intervals where these functions, along with others, decide to take a downward slope. We'll be focusing on the function f(x) = -5|x + 1| + 10 and figuring out where it's decreasing. Plus, we'll throw in another function (which you'll define) and find the common ground where both functions are losing altitude. Grab your graphing tools, and let's get started!

Understanding Absolute Value Functions

First off, let's break down the basics. Absolute value functions, at their core, give us the magnitude (or size) of a number, regardless of its sign. Think of it as the distance from zero. So, |x| is simply x if x is positive or zero, and -x if x is negative. This simple concept leads to some pretty cool graphs that look like V shapes. The general form we often see is f(x) = a|x - h| + k, where:

• a determines the direction and steepness of the V. If 'a' is positive, the V opens upwards; if it's negative, it opens downwards. The larger the absolute value of 'a', the steeper the V.
• (h, k) is the vertex, the pointy part of the V. It's where the action happens, where the function changes direction.

In our case, we have f(x) = -5|x + 1| + 10. Let's dissect this:

• -5 tells us the V opens downwards (because it's negative) and is quite steep (because 5 is a significant number).
• x + 1 can be rewritten as x - (-1), so h = -1. This means our V is shifted one unit to the left.
• +10 means our entire graph is lifted 10 units up.

So, the vertex of our function is at (-1, 10). This is a crucial point because it's where the function switches from increasing to decreasing (or vice versa). To the left of this point, our function will be climbing, and to the right, it'll be sliding down.

Graphing f(x) = -5|x + 1| + 10

Alright, time to put this all together and sketch the graph. We already know the vertex is at (-1, 10). Since the graph opens downwards, we can imagine a V shape pointing down from this peak. The '-5' in front of the absolute value isn't just about direction; it also affects the slope. A larger absolute value means a steeper slope. In this case, for every 1 unit we move away from the vertex horizontally, we move 5 units vertically.

Let's plot a couple more points to get a clearer picture:

• If x = 0, then f(0) = -5|0 + 1| + 10 = -5 + 10 = 5. So, we have a point at (0, 5).
• If x = -2, then f(-2) = -5|-2 + 1| + 10 = -5 + 10 = 5. So, we have another point at (-2, 5).

Now we can connect the dots! Starting from the vertex at (-1, 10), draw a line down and to the right through (0, 5) and another line down and to the left through (-2, 5). You've got your V shape, and that's the graph of f(x) = -5|x + 1| + 10.

Identifying the Decreasing Interval for f(x)

The question asks us where the function is decreasing. Remember, decreasing means the y-values are getting smaller as the x-values increase. Looking at our graph, this happens on the right side of the vertex. So, the function is decreasing for all x-values greater than -1. In interval notation, we write this as (-1, ∞). The parenthesis around -1 indicates that -1 itself is not included in the interval because that's where the function momentarily stops decreasing at the vertex.

Introducing Function 2 and Finding Common Decreasing Intervals

Now, let's spice things up. Imagine we have another function, let's call it g(x). For this example, let's consider g(x) = -x^2 - 2x + 3. This is a quadratic function, and its graph is a parabola opening downwards. To find where g(x) is decreasing, we need to think about its vertex and shape.

Analyzing g(x) = -x^2 - 2x + 3

This is a downward-facing parabola (because of the negative sign in front of the x² term). The x-coordinate of the vertex can be found using the formula x = -b / 2a, where a and b are the coefficients in the quadratic equation ax² + bx + c. In our case, a = -1 and b = -2, so the x-coordinate of the vertex is x = -(-2) / (2 * -1) = -1.

To find the y-coordinate of the vertex, we plug x = -1 back into g(x): g(-1) = -(-1)² - 2(-1) + 3 = -1 + 2 + 3 = 4. So, the vertex of g(x) is at (-1, 4).

Since the parabola opens downwards, it's increasing to the left of the vertex and decreasing to the right. Therefore, g(x) is decreasing on the interval (-1, ∞). This is the same decreasing interval as our absolute value function, f(x)!

Finding the Common Decreasing Interval

Here's the exciting part: we want to know where both functions are decreasing. We've already figured out that f(x) is decreasing on (-1, ∞) and g(x) is decreasing on (-1, ∞). So, the interval where they both decrease is simply the overlap of these two intervals. In this case, it's the same interval: (-1, ∞).

Representing the Interval on a Number Line

To represent this on a number line, draw a line and mark -1 on it. Since we're including all numbers greater than -1, we'll draw an open circle at -1 (to show that -1 itself is not included) and then shade the line to the right, indicating all the numbers up to infinity.

Visualizing with a Graphing Tool

For a more visual understanding, it's super helpful to use a graphing tool like Desmos or GeoGebra. You can plot both functions, f(x) = -5|x + 1| + 10 and g(x) = -x^2 - 2x + 3, and visually confirm where they are both decreasing. You'll see that to the right of x = -1, both graphs are heading downwards.

Wrapping Up

So, there you have it! We've navigated the world of absolute value functions, graphed them with confidence, and identified intervals where functions are decreasing. We even threw in another function and found the common ground where both were heading south. Remember, the key is understanding the transformations that affect the shape and position of the graph, and then carefully analyzing the intervals. Keep practicing, and you'll be a graphing guru in no time!

In this article, we explored how to graph the absolute value function f(x) = -5|x + 1| + 10 and determine its decreasing interval. We learned that the vertex of the function is a crucial point for understanding its behavior. By analyzing the coefficients and constants in the function, we identified the vertex at (-1, 10) and the downward-opening nature of the graph. This allowed us to pinpoint the decreasing interval as (-1, ∞).

We then introduced a second function, g(x) = -x^2 - 2x + 3, a quadratic function represented by a downward-facing parabola. By finding the vertex of the parabola at (-1, 4) and recognizing its shape, we determined that g(x) is also decreasing on the interval (-1, ∞). This led us to the conclusion that the common decreasing interval for both functions is also (-1, ∞).

Finally, we discussed how to represent this interval on a number line, using an open circle at -1 to indicate exclusion and shading the line to the right to represent all numbers greater than -1. We also highlighted the usefulness of graphing tools like Desmos or GeoGebra in visualizing the functions and their decreasing intervals.

This process demonstrates a powerful approach to analyzing functions and their behavior, combining algebraic analysis with graphical representation for a comprehensive understanding.