Transformations Of $f(x) = \sqrt{x}$: A Visual Guide

Hey guys! Today, we're diving deep into the fascinating world of function transformations, specifically focusing on our beloved square root function, $f(x) = \sqrt{x}$. We're going to take this basic function and put it through the wringer, reflecting it across the x-axis and then giving it a good old vertical stretch by a factor of 2. Our mission, should we choose to accept it, is to figure out what the resulting graph looks like. So grab your graphing calculators, your favorite colored pens, and let's get this party started!

Understanding the Base Function: $f(x) = \sqrt{x}$

Before we start bending and twisting our function, let's get reacquainted with the OG, the mother function: $f(x) = \sqrt{x}$. You know, the one that starts at the origin $(0,0)$ and gracefully curves upwards and to the right. Its domain is all non-negative real numbers ($x \ge 0$), and its range is also all non-negative real numbers ($y \ge 0$). This little guy is the foundation for many other functions, and understanding its basic shape and behavior is crucial for tackling transformations. Think of it as the blank canvas before the artist starts painting. We know its key points: $(0,0)$, $(1,1)$, $(4,2)$, $(9,3)$, and so on. These points are our anchors as we move on to the more exciting stuff.

Transformation 1: Reflection Across the x-axis

Alright, first up on our transformation agenda is reflecting $f(x) = \sqrt{x}$ across the x-axis. What does this even mean, you ask? Well, imagine the x-axis is a mirror. When you reflect a point across the x-axis, its x-coordinate stays the same, but its y-coordinate flips its sign. So, a point $(a, b)$ becomes $(a, -b)$. Applying this to our function, if $y = \sqrt{x}$, then reflecting across the x-axis means our new function, let's call it $g(x)$, will have outputs that are the negatives of the original outputs. Therefore, $g(x) = -f(x) = -\sqrt{x}$.

Let's look at our key points for $g(x) = -\sqrt{x}$:

• $(0,0)$ stays $(0,0)$ because $-0 = 0$.
• $(1,1)$ becomes $(1,-1)$.
• $(4,2)$ becomes $(4,-2)$.
• $(9,3)$ becomes $(9,-3)$.

See the pattern, guys? The graph of $g(x) = -\sqrt{x}$ looks like a mirror image of $f(x) = \sqrt{x}$ flipped upside down. Instead of curving upwards, it now curves downwards from the origin. The domain remains the same ($x \ge 0$), but the range is now all non-positive real numbers ($y \le 0$). It's like taking the original curve and pushing it through the x-axis.

Transformation 2: Vertical Stretch by a Factor of 2

Now, we take our reflected function, $g(x) = -\sqrt{x}$, and apply a vertical stretch by a factor of 2. What this means is that we're going to multiply the entire output of the function by 2. So, if a point on $g(x)$ is $(a, b)$, then the corresponding point on our new function, let's call it $h(x)$, will be $(a, 2b)$. We're essentially pulling the graph upwards (or downwards, in this case) away from the x-axis.

Applying this to $g(x) = -\sqrt{x}$, our final function $h(x)$ becomes $h(x) = 2 imes g(x) = 2 imes (-\sqrt{x}) = -2\sqrt{x}$.

Let's update our key points for $h(x) = -2\sqrt{x}$:

• $(0,0)$ remains $(0,0)$ because $2 imes 0 = 0$.
• $(1,-1)$ becomes $(1, 2 imes -1) = (1,-2)$.
• $(4,-2)$ becomes $(4, 2 imes -2) = (4,-4)$.
• $(9,-3)$ becomes $(9, 2 imes -3) = (9,-6)$.

So, what does this stretch do to the graph? Since we're multiplying the negative outputs by 2, the graph gets stretched vertically and pulled further away from the x-axis. The downward curve becomes steeper. The domain still remains $x \ge 0$, and the range is now all non-positive real numbers that are twice as far from zero, so $y \le 0$. It's like taking the already upside-down curve and elongating it, making it more dramatic.

Putting It All Together: The Resulting Graph

So, we started with $f(x) = \sqrt{x}$, reflected it across the x-axis to get $g(x) = -\sqrt{x}$, and then vertically stretched $g(x)$ by a factor of 2 to arrive at our final function, $h(x) = -2\sqrt{x}$. Let's recap the journey of our key points:

• Original $f(x) = \sqrt{x}$: $(0,0), (1,1), (4,2), (9,3)$
• After reflection $g(x) = -\sqrt{x}$: $(0,0), (1,-1), (4,-2), (9,-3)$
• After vertical stretch $h(x) = -2\sqrt{x}$: $(0,0), (1,-2), (4,-4), (9,-6)$

When you look at the graph of $h(x) = -2\sqrt{x}$, you'll see a curve that starts at the origin $(0,0)$ and extends downwards and to the right. It will be steeper than the graph of $y = -\sqrt{x}$ because of the vertical stretch. The domain is still $[0, infty)$ and the range is $(- infty, 0]$.

Selecting the Correct Graph

Now, when you're presented with a set of graphs, you need to look for the one that exhibits these characteristics. Here's what to check for:

1. Starting Point: Does the graph start at the origin $(0,0)$?
2. Direction: Does the curve extend downwards and to the right? If it extended upwards, it would be a stretch without a reflection across the x-axis, or if it went leftwards, it would involve horizontal transformations.
3. Steepness: Does the curve appear significantly steeper than a standard square root function reflected across the x-axis? You can test this by looking at specific points. For instance, does the graph pass through $(1, -2)$? Does it pass through $(4, -4)$? If it passes through these points, it's highly likely to be the correct graph.

Common Pitfalls to Avoid:

• Confusing Reflection and Stretch: Make sure you're not just looking at a graph that's vertically stretched but still going upwards, or a reflection that hasn't been stretched.
• Incorrect Axis of Reflection: Reflecting across the y-axis would produce a different result ($y = sqrt{-x}$).
• Horizontal Transformations: Ensure there are no horizontal shifts or stretches involved. The domain starting at $x=0$ is a key indicator.

By systematically applying these transformations and understanding how each affects the parent function $f(x) = \sqrt{x}$, you can confidently identify the correct graph. It's all about breaking down the problem into smaller, manageable steps. Remember, math is like building blocks; understand the foundation, and you can construct anything!

So, the graph that represents the function $f(x) = \sqrt{x}$ reflected across the x-axis and vertically stretched by a factor of 2 is the one that starts at $(0,0)$, goes downwards, and is steeper than the basic reflected square root function, specifically passing through points like $(1,-2)$ and $(4,-4)$. Keep these characteristics in mind, and you'll nail it every time, guys!