What's the Effect of a Reflection on a Graph (in the X-Axis or Y-Axis)?

A reflection in the x-axis changes every output from $y$ to (-y), while a reflection in the y-axis changes every input from $x$ to (-x). Therefore, $y=f(x)$ becomes $y=-f(x)$ in the x-axis and $y=f(-x)$ in the y-axis.

How the Reflections Work

A reflection in the x-axis reverses the vertical position of every point. The $x$-coordinate remains unchanged, but the $y$-coordinate changes sign.

A reflection in the y-axis reverses the horizontal position of every point. The $y$-coordinate remains unchanged, but the $x$-coordinate changes sign.

For example, consider

$$f(x)=(x-2)^2-1.$$

Its vertex is $(2,-1)$. After reflection in the x-axis,

$$y=-f(x)=-(x-2)^2+1,$$

so the vertex becomes $(2,1)$. After reflection in the y-axis,

$$y=f(-x)=(-x-2)^2-1=(x+2)^2-1,$$

so the vertex becomes $(-2,-1)$.

Points on the reflecting axis remain fixed. Therefore, x-intercepts stay in the same positions under reflection in the x-axis, while points on the y-axis stay fixed under reflection in the y-axis.

The same rules apply to every key feature. Under an x-axis reflection, a maximum becomes a minimum, vertical asymptotes remain in place, and the range changes sign. Under a y-axis reflection, left and right swap, horizontal asymptotes keep the same equation, and the domain is reflected.

For a quick point check, if $(3,4)$ lies on the original graph, then $(3,-4)$ lies on its x-axis reflection and $(-3,4)$ lies on its y-axis reflection. This verifies the algebra.

A common misconception is that reflection in the y-axis gives $y=-f(x)$. This is incorrect: a negative sign outside the function changes outputs and reflects in the x-axis; replacing $x$ with (-x) changes inputs and reflects in the y-axis.

IB Exam Technique

This is IB Mathematics AA SL 2.11: Transformation of functions and may be tested on Paper 1 or Paper 2. For a sketch, preserve the graph's shape and label transformed intercepts, vertices, and asymptotes; for algebraic questions, use brackets carefully when calculating $f(-x)$.