Reflecting And Shifting Logarithmic Graphs

Hey math lovers! Let's dive into the awesome world of logarithmic functions and transformations. Today, we're tackling a problem that involves reflecting a graph and shifting it. So, grab your pencils and let's get this done!

Understanding the Transformations

We're given the function $f(x) = \log_2 x$. Our mission, should we choose to accept it, is to find the function $g(x)$ that represents the graph of $f(x)$ after two transformations:

1. Reflection in the x-axis: When you reflect a function's graph across the x-axis, you essentially flip it upside down. Mathematically, this means changing the sign of the entire function. So, if you have $f(x)$, reflecting it in the x-axis gives you $-f(x)$. In our case, reflecting $f(x) = \log_2 x$ in the x-axis would result in $-\log_2 x$.

2. Shifting left by 2 units: Shifting a graph to the left by a certain number of units means we need to add that number to the input variable (x) inside the function. If we're shifting left by 2 units, we replace 'x' with '(x+2)'. So, if we had a function $h(x)$, shifting it left by 2 units would give us $h(x+2)$.

Now, the order of transformations can sometimes matter, but in this case, we're reflecting the original function first and then shifting the reflected function. Let's break it down step-by-step.

Step 1: Reflecting $f(x)$ in the x-axis

We start with our base function $f(x) = \log_2 x$. Reflecting this graph in the x-axis means we take the negative of the entire function. So, the new function, let's call it $h(x)$ for now, becomes:

$h(x) = -f(x) = -\log_2 x$

Think about what this does to the graph. The original $\log_2 x$ graph increases as x increases and approaches the y-axis asymptotically from the right. When we negate it, the graph will now decrease as x increases, and it will approach the y-axis asymptotically from the right, but from above. The key points change too. For instance, $(1, 0)$ on the original graph becomes $(1, 0)$ after reflection (since $-0=0$). However, $(2, 1)$ becomes $(2, -1)$, and $(1/2, -1)$ becomes $(1/2, 1)$. It's like looking at the original graph in a mirror placed on the x-axis.

Step 2: Shifting the reflected graph left by 2 units

Now, we take our reflected function $h(x) = -\log_2 x$ and shift its graph to the left by 2 units. To shift a function left by 'c' units, we replace 'x' with '(x+c)' in the function's expression. Here, c = 2. So, we replace 'x' in $h(x)$ with '(x+2)'.

This gives us our final function, $g(x)$:

$g(x) = h(x+2) = -\log_2(x+2)$

Let's visualize this shift. If we had a point $(a, b)$ on the graph of $h(x)$, shifting it left by 2 units moves it to $(a-2, b)$ on the graph of $g(x)$. For example, the point $(1, 0)$ on $h(x)$ (where $-\log_2 1 = 0$) would move to $(1-2, 0) = (-1, 0)$ on $g(x)$. The point $(2, -1)$ on $h(x)$ (where $-\log_2 2 = -1$) would move to $(2-2, -1) = (0, -1)$ on $g(x)$. The vertical asymptote, which was the y-axis (x=0) for $h(x)$, now shifts 2 units to the left, becoming the line $x = -2$ for $g(x)$. This makes sense because the logarithm function $\log_2 u$ is only defined for $u > 0$. In $g(x) = -\log_2(x+2)$, the argument of the logarithm is $(x+2)$. For the logarithm to be defined, we must have $x+2 > 0$, which means $x > -2$. This inequality $x > -2$ precisely describes the domain of the function $g(x)$, and its boundary, $x = -2$, is the vertical asymptote.

Analyzing the Options

Now, let's look at the given options to see which one matches our result:

• $g(x)=\log _2(-x-2)$: This function involves reflection across the y-axis (due to the negative inside the log) and a shift. If we factor out the -1 from inside the log, we get $\log_2(-(x+2))$. This represents a reflection across the y-axis first, then a shift left by 2. That's not what we did.
• $g(x)=-\log _2 x-2$: This function reflects the original graph in the x-axis (giving $-\log_2 x$) and then shifts the entire resulting graph down by 2 units. The '-2' is outside the logarithm, indicating a vertical shift, not a horizontal one.
• $g(x)=-\log _2(x+2)$: This matches our derived function! It correctly applies the reflection in the x-axis ($-\log_2 x$) and then shifts the result left by 2 units by replacing $x$ with $(x+2)$.
• $g(x)=2-\log _2 x$: This can be rewritten as $g(x) = -\log_2 x + 2$. This reflects the original graph in the x-axis and then shifts the entire resulting graph up by 2 units. Again, the '+2' indicates a vertical shift.

Conclusion

So, the function that represents reflecting the graph of $f(x)=\log_2 x$ in the x-axis and shifting left by 2 units is $g(x)=-\log _2(x+2)$.

Understanding these transformations is super handy. Remember:

• Reflection across the x-axis: $f(x) \rightarrow -f(x)$
• Reflection across the y-axis: $f(x) \rightarrow f(-x)$
• Shift left by 'c' units: $f(x) \rightarrow f(x+c)$
• Shift right by 'c' units: $f(x) \rightarrow f(x-c)$
• Shift up by 'c' units: $f(x) \rightarrow f(x)+c$
• Shift down by 'c' units: $f(x) \rightarrow f(x)-c$

Keep practicing these transformations, guys, and you'll become a master of graph manipulation in no time! Let me know if you have any questions or want to try another one!