Reflecting Exponential Functions: Find The Point On G(x)

Hey guys! Today, we're diving into the cool world of exponential functions and how reflections can change them. Specifically, we're looking at the function f(x)=rac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^x. We're going to reflect this bad boy across the y-axis to create a new function, $g(x)$. Then, the big question is: which of the given ordered pairs actually sits on this new function $g(x)$? Let's break it down!

Understanding Reflections Across the y-axis

So, when we talk about reflecting a function across the y-axis, what does that actually mean mathematically? Imagine you have a graph, and you place a mirror right on the y-axis. The reflection is what you'd see on the other side of that mirror. For any given point $(x, y)$ on the original function, its reflection across the y-axis will be at $(-x, y)$. Notice that the y-coordinate stays the same, but the x-coordinate flips its sign. This is the key transformation we need to apply to our function $f(x)$ to get $g(x)$.

If $f(x)$ is our original function, then reflecting it across the y-axis to create $g(x)$ means that $g(x) = f(-x)$. So, wherever we see an 'x' in the original function $f(x)$, we're going to replace it with '-x' to find the equation for $g(x)$. This simple substitution is the magic that performs the reflection. It's like time travel for the x-values – they go backward! This concept is fundamental when dealing with transformations of functions, and understanding it will help you tackle all sorts of problems involving shifts, stretches, and flips of graphs. So, let's get ready to apply this rule to our specific function and see what we discover. It's going to be an awesome ride through the land of math!

Applying the Reflection to f(x)

Alright, let's get down to business with our specific function: f(x)=\\\frac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^x. To find the equation for $g(x)$, which is the reflection of $f(x)$ across the y-axis, we apply the rule $g(x) = f(-x)$. This means we substitute '-x' for every 'x' in the expression for $f(x)$.

So, g(x) = f(-x) = \\\rac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^{-x}.

Now, dealing with a negative exponent can be a bit tricky, but we know how to handle it. Remember that for any non-zero number 'a' and any real number 'n', $a^{-n} = \\\rac{1}{a^n}$. Also, a related property is \\\left(\\,\rac{a}{b}\\,\ ight)^{-n} = \\\rac{b^n}{a^n} = \\\ ight(\\,\rac{b}{a}\\,\ ight)^n. This is super handy!

Applying this rule to our $g(x)$, we get:

g(x) = \\\rac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^{-x} = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{x}.

So, our reflected function is g(x) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{x}. Keep this equation handy because we'll use it to check our potential ordered pairs. It's pretty neat how a simple sign change in the exponent transforms the base of the exponential function, right? This is a core concept in understanding how exponential functions behave under transformations. The reciprocal of the base arises directly from the property of negative exponents, making the reflection across the y-axis a powerful tool for altering the growth or decay behavior of the function.

Testing the Ordered Pairs

Now that we have the equation for $g(x)$, which is g(x) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{x}, we need to figure out which of the given ordered pairs lies on this function. An ordered pair $(x, y)$ is on the function $g(x)$ if, when you plug the x-value into $g(x)$, you get the corresponding y-value. Let's test each option one by one.

Option A: $\\\left(\\,`-2, \rac{25}{24}\\\right)$**

Here, $x = -2$. Let's plug this into our $g(x)$ function:

g(-2) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{-2}

Using our exponent rule \\\left(\\,\rac{a}{b}\\,\ ight)^{-n} = \\\ ight(\\,\rac{b}{a}\\,\ ight)^n, we get:

g(-2) = \\\rac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^{2}

Now, we square the fraction:

g(-2) = \\\rac{1}{6}\\\left(\\,\rac{2^2}{5^2}\\,\ ight) = \\\rac{1}{6}\\\left(\\,\rac{4}{25}\\,\ ight)

Finally, multiply the fractions:

$g(-2) = \\\rac{1 \\times 4}{6 \\times 25} = \\\rac{4}{150}$

We can simplify this fraction by dividing both the numerator and the denominator by 2:

$g(-2) = \\\rac{2}{75}$

Our calculated y-value is $\\\frac{2}{75}$, but the given y-value in option A is $\\\frac{25}{24}$. Since $\\\frac{2}{75} \\neq \\\rac{25}{24}$, option A is not on the graph of $g(x)$.

Option B: $\\\left(\\,`-3, \rac{4}{375}\\\right)$**

Here, $x = -3$. Let's substitute this into $g(x)$:

g(-3) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{-3}

Again, we use the negative exponent rule:

g(-3) = \\\rac{1}{6}\\\left(\\,\rac{2}{5}\\,\ ight)^{3}

Now, we cube the fraction:

g(-3) = \\\rac{1}{6}\\\left(\\,\rac{2^3}{5^3}\\,\ ight) = \\\rac{1}{6}\\\left(\\,\rac{8}{125}\\,\ ight)

Multiply the fractions:

$g(-3) = \\\rac{1 \\times 8}{6 \\times 125} = \\\rac{8}{750}$

Let's simplify this fraction. We can divide both the numerator and the denominator by 2:

$g(-3) = \\\rac{4}{375}$

Look at that! The calculated y-value is $\\\frac{4}{375}$, which is exactly the y-value given in option B. Therefore, the ordered pair $\\\left(\\,`-3, \rac{4}{375}\\\right)$ is on the graph of $g(x)$. We found our answer, but let's check the remaining options just to be thorough and solidify our understanding.

Option C: $\\\left(2, \rac{3}{25}\\\right)$**

Here, $x = 2$. Let's plug this into our $g(x)$ function:

g(2) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{2}

First, square the fraction:

g(2) = \\\rac{1}{6}\\\left(\\,\rac{5^2}{2^2}\\,\ ight) = \\\rac{1}{6}\\\left(\\,\rac{25}{4}\\,\ ight)

Now, multiply the fractions:

$g(2) = \\\rac{1 \\times 25}{6 \\times 4} = \\\rac{25}{24}$

Our calculated y-value is $\\\frac{25}{24}$, but the given y-value in option C is $\\\frac{3}{25}$. Since $\\\frac{25}{24} \\neq \\\rac{3}{25}$, option C is not on the graph of $g(x)$.

Option D: $\\\left(3, \rac{27}{1000}\\\right)$**

Here, $x = 3$. Let's substitute this into $g(x)$:

g(3) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{3}

Cube the fraction:

g(3) = \\\rac{1}{6}\\\left(\\,\rac{5^3}{2^3}\\,\ ight) = \\\rac{1}{6}\\\left(\\,\rac{125}{8}\\,\ ight)

Multiply the fractions:

$g(3) = \\\rac{1 \\times 125}{6 \\times 8} = \\\rac{125}{48}$

Our calculated y-value is $\\\frac{125}{48}$, but the given y-value in option D is $\\\frac{27}{1000}$. Since $\\\frac{125}{48} \\neq \\\rac{27}{1000}$, option D is not on the graph of $g(x)$.

Conclusion

After carefully testing each ordered pair with our derived function g(x) = \\\rac{1}{6}\\\left(\\,\rac{5}{2}\\,\ ight)^{x}, we found that only one pair satisfies the equation. The key to solving this problem lies in understanding how a reflection across the y-axis transforms a function, specifically by replacing $x$ with $-x$. We then used the properties of exponents, particularly those involving negative exponents, to simplify the expression for $g(x)$.

By substituting the x-values from each ordered pair into $g(x)$ and comparing the resulting y-values with the given y-values, we were able to determine which point lies on the graph. It's crucial to perform these calculations accurately, paying close attention to the order of operations and exponent rules. The ability to manipulate exponential expressions and understand function transformations is a cornerstone of algebra and is essential for success in higher-level mathematics. So, next time you see a reflection problem, remember these steps, and you'll be good to go! Keep practicing, guys, and you'll master these concepts in no time. The ordered pair that is on $g(x)$ is B. $\\\left(\\,`-3, \rac{4}{375}\\\right)$.