Why Symmetry Helps with Sketching
A graph or contour map can contain repeated information. When a function has a symmetry, reflecting part of its graph or a level curve produces another part that must also be present. Recognizing this lets you focus on one side or one family of traces, then use the symmetry to complete the picture.
The important step is to identify exactly which transformation leaves the function unchanged. Being even in \(x\), being even in \(y\), and being unchanged when \(x\) and \(y\) are exchanged are different properties. They imply different reflections and different relationships between traces.
Definitions and Domain Conditions
Let \(D\) be the domain of a function \(f:D\to\mathbb{R}\). A reflection or exchange can only be a symmetry of the function if the transformed input is also in its domain. Thus, the relevant domain must be preserved by that transformation.
For example, if \(D=\mathbb{R}^2\), all the transformed points mentioned in these definitions are in the domain. On a smaller domain, the condition must be checked: a formula alone cannot guarantee a symmetry if the domain removes one of the reflected or exchanged points.
The graph of \(f\) is the set of points \((x,y,z)\) with \((x,y)\in D\) and \(z=f(x,y)\). A vertical trace at \(y=b\) is the part of the graph obtained by fixing \(y=b\); its height as a function of \(x\) is \(z=f(x,b)\). At \(x=a\), the corresponding trace has height \(z=f(a,y)\). A level curve at height \(c\) is the set of domain points satisfying \(f(x,y)=c\).
What Each Symmetry Does
Evenness in \(x\) pairs inputs \((x,y)\) and \((-x,y)\) with the same height. In the graph, this is reflection across the plane \(x=0\). It makes each trace at a fixed \(y=b\) symmetric across \(x=0\), and makes every level curve symmetric across the \(y\)-axis in the \(xy\)-plane.
Evenness in \(y\) instead pairs \((x,y)\) with \((x,-y)\). It reflects the graph across the plane \(y=0\), makes each fixed-\(x\) trace symmetric across \(y=0\), and reflects each level curve across the \(x\)-axis.
Exchanging \(x\) and \(y\) has a different effect. The graph is reflected across the vertical plane \(x=y\), and each level curve is reflected across the line \(x=y\) in the \(xy\)-plane. A trace at \(x=a\) corresponds to a trace at \(y=a\); these are generally in different vertical planes, rather than being a trace reflected within the same plane.
Why the Symmetry Rules Hold
The claims follow directly from the definition of a graph and from the equalities defining each kind of symmetry. The following proof makes clear both the geometric transformation and the domain requirement.
Suppose \(f(-x,y)=f(x,y)\). If \((x,y,z)\) is on the graph, then \(z=f(x,y)\). The reflected input \((-x,y)\) is in the domain, and \(f(-x,y)=z\). Therefore \((-x,y,z)\) is also on the graph. Reflecting twice returns \((x,y,z)\), so the graph is symmetric across \(x=0\).
If \(z=f(x,y)\) and \(f(x,-y)=f(x,y)\), then \(z=f(x,-y)\). Hence \((x,-y,z)\) is on the graph whenever \((x,y,z)\) is. This is reflection across \(y=0\).
If \(z=f(x,y)\) and \(f(y,x)=f(x,y)\), then \(z=f(y,x)\). Thus \((y,x,z)\) is on the graph whenever \((x,y,z)\) is. This is reflection across \(x=y\).
The same equalities prove the level-curve claims. For instance, if \(f(x,y)=c\) and \(f(-x,y)=f(x,y)\), then \(f(-x,y)=c\) as well. Thus reflecting any point on the level curve across the \(y\)-axis produces another point on that same level curve. Replacing \((-x,y)\) by \((x,-y)\) or \((y,x)\) proves the corresponding claims for evenness in \(y\) and exchange symmetry.
For traces, fix one coordinate before applying the equality. If \(f\) is even in \(x\), then at a fixed \(y=b\), \(f(-x,b)=f(x,b)\), so the two points of the trace at \(x\) and \(-x\) have the same height. Under exchange symmetry, \(f(a,y)=f(y,a)\), which says that the height at \((a,y)\) matches the height at \((y,a)\); this pairs the trace at \(x=a\) with the trace at \(y=a\).
Worked Example: Evenness in One Variable
Worked Example: Evenness in One Variable
Consider \(f(x,y)=x^2+y\) on \(\mathbb{R}^2\). Test for evenness in each variable, then describe the relevant traces and level curves.
To test evenness in \(x\), substitute \(-x\):
Thus \(f\) is even in \(x\). But substituting \(-y\) gives
which is not generally equal to \(x^2+y\). For example, at \((x,y)=(1,2)\), \(f(1,2)=1+2=3\), while \(f(1,-2)=1-2=-1\). So \(f\) is not even in \(y\).
Fix \(y=2\). The trace has height \(z=x^2+2\). At \(x=1\) and \(x=-1\), the heights are \(1+2=3\) and \((-1)^2+2=3\), respectively. More generally, replacing \(x\) with \(-x\) leaves this trace unchanged.
For a level \(c=3\), solve \(f(x,y)=3\):
The point \((1,2)\) lies on this level curve because \(1^2+2=3\). Its reflection \((-1,2)\) also lies on it because \((-1)^2+2=3\). This verifies the predicted symmetry across the \(y\)-axis.
Worked Example: Evenness in Both Variables
Worked Example: Evenness in Both Variables
Let \(g(x,y)=x^2+4y^2\) on \(\mathbb{R}^2\). Determine its reflection symmetries and examine one vertical trace and one level curve.
Reflecting in \(x\) gives
Reflecting in \(y\) gives
Therefore the graph is symmetric across both \(x=0\) and \(y=0\), and its level curves are symmetric across both coordinate axes. These two equalities do not imply symmetry under exchanging \(x\) and \(y\): in fact \(g(1,0)=1\), whereas \(g(0,1)=4\).
At fixed \(y=1\), the trace has height \(z=x^2+4\). At \(x=2\) its height is \(2^2+4=8\); at \(x=-2\) it is \((-2)^2+4=8\), as the \(x\)-evenness predicts.
The level \(c=4\) is given by
For example, \((0,1)\) is on it because \(0^2+4(1)^2=4\). Its reflection \((0,-1)\) is also on it because \(0^2+4(-1)^2=4\). Likewise, \((2,0)\) and \((-2,0)\) both give value \(4\). These substitutions check the predicted axis reflections on the level curve.
Worked Example: Symmetry Under Exchanging Variables
Worked Example: Symmetry Under Exchanging Variables
Let \(h(x,y)=x^2+xy+y^2\) on \(\mathbb{R}^2\). Find the exchange symmetry and use it to relate two traces and verify points on a level curve.
Exchange the variables:
Thus the graph is symmetric across the plane \(x=y\), and each level curve is symmetric across the line \(x=y\). At fixed \(x=1\), the trace has height
The corresponding fixed-\(y=1\) trace has height \(h(x,1)=x^2+x+1\). At \(y=2\) on the first trace, the height is \(1+2+2^2=7\). At \(x=2\) on the second, it is \(2^2+2+1=7\). The equal heights verify the paired points \((1,2,7)\) and \((2,1,7)\).
For level \(c=3\), the point \((1,1)\) is on the level curve since \(1+1+1=3\). It remains fixed under exchange. Another point is \((1,0)\), where \(h(1,0)=1+0+0=1\), so it is not on the level \(3\). To check a non-fixed pair on that level, take \((\sqrt{3},0)\): \(h(\sqrt{3},0)=3\). Its exchanged point \((0,\sqrt{3})\) also gives \(h(0,\sqrt{3})=3\). Thus both points lie on the level curve, as required. The same verification works for any point on the level: exchanging its coordinates leaves the value unchanged.
Using Symmetry to Reduce Sketching Work
Symmetry reduces work by letting you construct matching pieces from a representative piece. If a graph is even in \(x\), sketch the trace or contour on one side of \(x=0\), then reflect it across that plane or axis. If it is even in both variables, one quadrant of a level curve can determine its reflected pieces in the other three quadrants. With exchange symmetry, a part on one side of \(x=y\) determines its mirror on the other side.
For traces, distinguish a reflection from a match between different slices. Evenness in \(x\) says the trace at a fixed \(y=b\) mirrors within the plane \(y=b\). Exchange symmetry says the trace at \(x=a\) has matching heights to the trace at \(y=a\), after the coordinates are exchanged. It does not say that every trace in one family is identical to every trace in the other; the matching value \(a\) matters.
A Common Mistake: Confusing Exchange with Sign Change
Exchange symmetry is not the same as evenness. For \(h(x,y)=x^2+xy+y^2\), it would be incorrect to claim that \(h\) is even in \(x\) just because \(h(y,x)=h(x,y)\). A mistaken check might replace \((-1,2)\) by the exchanged point \((2,-1)\), then treat that as a comparison with \((1,2)\). But exchange changes \((x,y)\) to \((y,x)\); it does not change \(x\) to \(-x\).
The actual values show the distinction:
Since these values differ, \(h\) is not even in \(x\). The correction is to test each property with its own transformation: compare \(f(x,y)\) with \(f(-x,y)\) for evenness in \(x\), with \(f(x,-y)\) for evenness in \(y\), and with \(f(y,x)\) for exchange symmetry.
Check Your Understanding
Use the definitions and symmetry rules to answer the following questions.
- What domain condition is needed before claiming that \(f\) is even in \(x\)?
- If \(f\) is even in \(y\), which reflection leaves its graph unchanged, and which axis reflection applies to its level curves?
- For a function symmetric under exchanging \(x\) and \(y\), how does the trace at \(x=a\) relate to the trace at \(y=a\)?
- Is \(p(x,y)=x^2+y^2\) symmetric under exchange? Is \(q(x,y)=x^2+2y^2\)? Verify by substitution.
- Explain why exchange symmetry alone does not imply evenness in \(x\).