I keep trying to fold graphs in my head like paper snowflakes, but the symmetry keeps slipping away when the function is shifted. I know the basic tests: even if f(-x) = f(x), odd if f(-x) = -f(x). But what happens when the graph takes a little walk to the right or up? For example, g(x) = (x – 3)^2 + 2 looks like it should be mirror-y around x = 3, but the usual f(-x) check doesn’t match g(x), so my brain yells “not even!” and then the picture says “yes, but kind of.” Same confusion with q(x) = (x + 1)^3 + 5 – does the odd symmetry move to a new center, like around a point (a, b)? How do I actually test that with plugging in values? I feel like I’m supposed to compare x with something like 2a – x for a vertical mirror or check a centered version for point symmetry, but I don’t know the right way to write it. Could someone explain the plug-in tests for symmetry after translations (and maybe scalings), and how to spot the correct axis or center just from the formula, without graphing? A simple rule I can apply to examples like g(x) = (x – 3)^2 + 2 and q(x) = (x + 1)^3 + 5 would help a ton.
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3 Responses
I used to trip over this too-my “is it even/odd?” reflex kept shouting at the origin while the graph had quietly moved house. The fix is to move the test along with the graph. For mirror symmetry about a vertical line x = a, plug in symmetrically around a: g(a + h) = g(a − h) for all h (equivalently, g(2a − x) = g(x) for all x). For half-turn (odd-type) symmetry about a point (a, b), center the outputs too: g(a + h) − b = −(g(a − h) − b) for all h (equivalently, g(2a − x) = 2b − g(x)). A handy way to spot a and b from the formula: try to rewrite your function as g(x) = b + F(k(x − a)). Then the “core” F decides the symmetry. If F is even (like u^2, |u|, cos u), you get mirror symmetry about x = a. If F is odd (like u^3, u, sin u, tan u), you get point symmetry about (a, b). The stretch factors k (horizontal) and any outside multiplier c (vertical) don’t break the symmetry; they just scale it.
Now your examples. For g(x) = (x − 3)^2 + 2, write g(x) = 2 + F(x − 3) with F(u) = u^2 (even). Test: g(3 + h) = h^2 + 2 and g(3 − h) = (−h)^2 + 2 = h^2 + 2, so g(3 + h) = g(3 − h). That’s mirror symmetry about x = 3. In the “2a − x” form: a = 3, so g(6 − x) = g(x). For q(x) = (x + 1)^3 + 5, rewrite as q(x) = 5 + F(x − (−1)) with F(u) = u^3 (odd). Test: q(−1 + h) − 5 = h^3 and q(−1 − h) − 5 = (−h)^3 = −h^3, so q(−1 + h) − 5 = −(q(−1 − h) − 5). That’s point symmetry about (a, b) = (−1, 5). In the “2a − x” form: q(2(−1) − x) = q(−2 − x) = 2·5 − q(x) = 10 − q(x). If you ever forget, just shift and center: set H(h) = g(a + h) − b; H is even for mirror symmetry and odd for point symmetry.
Rule of thumb I keep taped to my mental desk: find the inside shift a from x − a (or x + 1 means a = −1), peek for any outside +b, and treat what’s left as F. If F is even, check g(a + h) = g(a − h). If F is odd, check g(a + h) + g(a − h) = 2b (same as g(2a − x) = 2b − g(x)). A quick refresher on even/odd functions (at the origin) is here, and everything above is just those tests after shifting and scaling: https://www.khanacademy.org/math/algebra2/polynomial-functions/even-and-odd-functions/a/even-and-odd-functions
Think of “folding” as moving your crease: mirror symmetry about x = a if g(a+h) = g(a−h) (i.e., g(2a−x) = g(x)), and point symmetry about (a, b) if g(a+h)−b = −(g(a−h)−b) (i.e., g(2a−x) = 2b−g(x)); you spot a and b by rewriting g(x) = F(k(x−a)) + b where F is even (mirror) or odd (point)-scalings don’t break the symmetry. For your examples: g(x) = (x−3)^2+2 = F(x−3)+2 with F(t)=t^2 even, so g(6−x)=g(x) (mirror about x=3), and q(x) = (x+1)^3+5 = F(x−(−1))+5 with F(t)=t^3 odd, so q(−2−x)=2·5−q(x) (point symmetry about (−1, 5)).
Totally feel you on the “paper snowflake” brain-folding! The clean way to test symmetry after translations is to shift your viewpoint to the suspected center/axis. For a vertical mirror about x = a, plug x = a + t and x = a − t and see if g(a + t) = g(a − t) for all t. Equivalently, you can check g(x) = g(2a − x). For point (odd-type) symmetry about a center (a, b), subtract off the height b and do the same trick: g(a + t) − b = −(g(a − t) − b). A quick repackage of that is g(a + t) + g(a − t) = 2b. I sometimes also say “for mirrors, g(a + t) + g(a − t) = 2g(a),” which feels intuitively right to me, though I admit I get tangled there because that sum actually depends on t in general, so I usually fall back to the plain equality test g(a + t) = g(a − t).
Now your examples snap into place. For g(x) = (x − 3)^2 + 2, try a = 3: g(3 + t) = t^2 + 2 and g(3 − t) = t^2 + 2, so it’s mirror-symmetric about x = 3. For q(x) = (x + 1)^3 + 5, try center (a, b) = (−1, 5): q(−1 + t) − 5 = t^3 and q(−1 − t) − 5 = −t^3, so it has point symmetry about (−1, 5). A handy “spot it from the formula” rule: if you can rewrite your function as A·F(B(x − a)) + b, then if F is even, you get mirror symmetry about x = a; if F is odd, you get point symmetry about (a, b). Horizontal or vertical scalings A and B keep that symmetry, though I sometimes trip over the sign of B and think it slides the axis to a/k (which isn’t right-best to just plug with a ± t and check). Hope this helps!