Reflection Across the x-axis: y = -f(x)

Reflecting a graph across the x-axis means flipping it upside down, as if the x-axis were a mirror lying flat on the page. Every point above the x-axis lands the same distance below it, and every point below the x-axis lands the same distance above it. The x-axis itself acts as the line of symmetry between the original graph and its reflection.

This is one of the core function transformations, alongside stretches, shifts, and the related y axis reflection, which flips a graph left-to-right instead of top-to-bottom. Learning to recognize the difference between the two is one of the most useful skills for graphing quickly without a table of values.

If \(y = f(x)\) is the original function, the reflection of that graph across the x-axis is given by:

\(y = -f(x)\)

In words: take the original function and multiply the entire output by \(-1\). Nothing happens to the \(x\)-value; only the sign of the \(y\)-value changes. So for any point \((x, y)\) on the graph of \(f(x)\), the point \((x, -y)\) lies on the graph of \(-f(x)\).

This is very different from replacing \(x\) with \(-x\). Writing \(f(-x)\) negates the input and reflects the graph across the y-axis, while \(-f(x)\) negates the output and reflects the graph across the x-axis. Mixing these two up is the single most common error students make with this topic.

1. Start with the equation \(y = f(x)\).
2. Multiply the entire right-hand side by \(-1\) to get \(y = -f(x)\).
3. Simplify the new equation if needed.
4. If you are graphing point by point, keep every \(x\)-coordinate the same and flip the sign of every \(y\)-coordinate.
5. Check that any \(x\)-intercepts (where \(y = 0\)) have not moved, since \(-0 = 0\).

Let \(f(x) = x^2 - 2x - 3\). This parabola opens upward with a minimum point (vertex) at \(x = 1\) and x-intercepts at \(x = -1\) and \(x = 3\).

To reflect this across the x-axis, apply the rule \(y = -f(x)\):

\(-f(x) = -(x^2 - 2x - 3) = -x^2 + 2x + 3\)

Notice that the x-intercepts \(x = -1\) and \(x = 3\) are unchanged, since \(y = 0\) there in both graphs. The vertex, however, moved from a minimum at \(x = 1\) to a maximum at \(x = 1\), because its \(y\)-value simply switched sign.

Now let \(g(x) = |x - 1|\), a V-shaped graph with its vertex (a minimum) at \(x = 1\).

Reflecting across the x-axis gives \(y = -g(x) = -|x - 1|\), which turns the upward V into a downward V with a maximum at the same \(x\)-value:

Again, only the sign of each output flipped. The location of the vertex along the x-axis never moved.

Because the rule only touches the output of the function, the how to write domain and range lesson is worth reviewing alongside this one: the domain of \(-f(x)\) is identical to the domain of \(f(x)\), since the same \(x\)-values are still allowed. The range, however, flips: if the original range is \(y \ge 0\), the reflected range becomes \(y \le 0\), and vice versa. Any x-intercepts are preserved exactly, while any maximum becomes a minimum and any minimum becomes a maximum.

Reflections rarely show up alone on a test. You will often see them combined with a horizontal translation or a vertical shift in the same equation, such as \(y = -f(x - 2) + 3\). When several transformations appear together, apply them in order: handle what happens inside the function first (shifts to \(x\)), then apply the reflection and any outside shifts to the result.

• Confusing \(-f(x)\) with \(f(-x)\): the first flips the graph vertically (across the x-axis), the second flips it horizontally (across the y-axis).
• Forgetting to distribute the negative sign across every term, especially when \(f(x)\) is a sum of several terms.
• Moving the x-intercepts by mistake; a correctly reflected graph always crosses the x-axis at the same spots as the original.
• Assuming the domain changes; it does not, since reflection across the x-axis only affects \(y\)-values.

A fast way to check a reflection is to pick two or three easy points from the original graph, flip the sign of each \(y\)-value, and plot the new points. If they trace a mirror image across the x-axis, the equation \(y = -f(x)\) is correct.

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