In the function , is a constant. The graph of is reflected across the -axis and then translated 8 units up to produce the graph of . Which equation defines ?

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SAT Nonlinear Functions Question 134 | Free SAT Prep | Aniko

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Question 134·Medium·Nonlinear Functions

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$f(x)=\dfrac{m}{(x+2)^2}-6$

In the function $f$ , $m$ is a constant. The graph of $y=f(x)$ is reflected across the $y$ -axis and then translated 8 units up to produce the graph of $y=g(x)$ . Which equation defines $g$ ?

For function transformation questions, remember the key rules: reflecting across the y-axis means replace $x$ with $-x$ in the function; vertical shifts are done by adding or subtracting a constant outside the function. Apply the transformations in the given order, simplify any constants carefully (e.g., $-6+8=2$ ), and then match the final algebraic form—especially the parts inside parentheses and the final constant term—to the answer choices.

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Think about reflections across the y-axis

When a graph is reflected across the $y$ -axis, how do the $x$ -coordinates of points change? How can you represent that change by modifying the function $f(x)$ ?

Write the equation after reflection

Take $f(x)=\dfrac{m}{(x+2)^2}-6$ and replace $x$ with $-x$ everywhere. Write down the new function $f(-x)$ .

Use the rule for vertical shifts

For a vertical translation up by 8 units, what number should you add or subtract from the entire right-hand side of the equation? Apply that to your expression from the reflection step.

Desmos Guide

Graph the original function

In Desmos, type y = 1/(x+2)^2 - 6 (you can use 1 for $m$ since the transformations we care about do not depend on the specific value of $m$ ). Observe its shape and its vertical asymptote at $x=-2$ and horizontal asymptote at $y=-6$ .

Graph the reflection across the y-axis

Now type y = 1/(-x+2)^2 - 6. Check that this new graph is a mirror image of the original across the $y$ -axis and that its vertical asymptote has moved to $x=2$ while the horizontal asymptote stays at $y=-6$ .

Graph the vertical shift and match to an answer choice

Type y = 1/(-x+2)^2 + 2. This is the result of shifting the reflected graph up by 8 units (its horizontal asymptote should now be at $y=2$ ). Compare this equation’s structure to the answer choices and select the one that matches it, just replacing the 1 with $m$ .

Step-by-step Explanation

Interpret the transformations

The graph of $y=f(x)$ is first reflected across the $y$ -axis, then moved (translated) 8 units up. We must apply these transformations to the equation of $f(x)$ in that order.

Reflect across the y-axis

To reflect a graph across the $y$ -axis, replace $x$ with $-x$ in the function rule.

Start with $f(x)=\dfrac{m}{(x+2)^2}-6$ and compute $f(-x)$ :

Replace every $x$ with $-x$ : $f(-x)=\dfrac{m}{(-x+2)^2}-6$ .

This is the equation of the reflected graph.

Translate 8 units up

Translating a graph up by 8 units adds 8 to the entire function value. That is, $y=f(-x)$ becomes $y=f(-x)+8$ .

So from $f(-x)=\dfrac{m}{(-x+2)^2}-6$ , we add 8:

f(-x)+8=\dfrac{m}{(-x+2)^2}-6+8=\dfrac{m}{(-x+2)^2}+2

This new function is $g(x)$ , so the correct equation is $g(x)=\dfrac{m}{(-x+2)^2}+2$ . This matches answer choice C.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation