I’m trying to wrap my head around symmetry of function graphs, and I think I’m mixing myself up. I’m fine with the usual even/odd thing: even if f(-x) = f(x), odd if f(-x) = -f(x). That makes sense for the y-axis and the origin.
Where I get stuck is when the graph is symmetric about some vertical line that isn’t the y-axis, like x = 2. For example, h(x) = (x – 2)^2. If I do the usual even test, h(-x) = (-x – 2)^2 = (x + 2)^2, which isn’t the same as (x – 2)^2, so it’s not even. But the parabola is clearly a mirror image around x = 2. I feel like I should be replacing x with something like 2 – x or maybe -(x – 2), but my brain keeps flipping left and right.
My (probably half-baked) attempt: I tried checking h(2 – x) against h(x). For this function, h(2 – x) = (2 – x – 2)^2 = (-x)^2 = x^2, which doesn’t match (x – 2)^2, so I guess that comparison isn’t the right one.
Simple number check that makes me think there’s symmetry: h(1) = (1 – 2)^2 = 1 and h(3) = (3 – 2)^2 = 1. So 1 and 3 are equal distances from 2 and give the same value. That feels like the correct intuition, but I don’t know how to turn it into a clean algebra test.
My questions:
– What’s the correct way (algebraically) to test if a function is symmetric about x = a, like x = 2, without graphing? Is something like “h(2 + t) equals h(2 – t)” the right idea, or am I holding the mirror in the wrong place?
– Bonus confusion: does it make sense to talk about an “odd” type of symmetry around a vertical line or around a point (like symmetry about (a, b)) for a function, and if so, what substitution would you check?
Thanks! I’m probably overthinking this, but I’d love a simple rule I can apply from the formula.
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3 Responses
Your mirror didn’t disappear. It just slid to a new address. When the mirror line is x = a, the reflection sends x to 2a − x. That’s the whole trick. Once you recenter your thinking at a, the usual even/odd tests come back to life.
How to test symmetry about a vertical line x = a
– Geometric reflection across x = a sends any point (x, y) to (2a − x, y).
– Algebraic test for vertical mirror symmetry: f(x) = f(2a − x) for every x in the domain for which both sides are defined.
– Same idea, written with a recentering: define F(t) = f(a + t). Then the graph is symmetric about x = a exactly when F is even, i.e., F(−t) = F(t). This is the “move the origin to a” view.
Your example, h(x) = (x − 2)²
– Use the reflection test with a = 2:
h(2·2 − x) = h(4 − x) = (4 − x − 2)² = (2 − x)² = (x − 2)² = h(x).
So yes, it’s symmetric about x = 2.
– Or use the recentered even test: set F(t) = h(2 + t) = (t)².
Then F(−t) = (−t)² = t² = F(t). Even after the shift, just as expected.
– Why h(2 − x) wasn’t the right comparison in your attempt: you changed x to 2 − x, but you didn’t compare symmetric points around 2. The right comparison is “plug in equal offsets on both sides,” i.e., check h(2 + t) versus h(2 − t) for an arbitrary t, or equivalently check h(x) versus h(2a − x).
A clean recipe you can use every time
1) Want vertical-line symmetry about x = a?
Check f(x) = f(2a − x) for all x.
Equivalently, let F(t) = f(a + t) and check F(−t) = F(t).
2) Want the “odd-style” symmetry, but not about the origin?
That naturally means 180° rotational symmetry about some point (a, b) (also called point symmetry).
The rotation by 180° about (a, b) sends (x, y) to (2a − x, 2b − y).
Algebraic test: f(2a − x) = 2b − f(x), for all x.
Equivalently, define G(t) = f(a + t) − b and check G(−t) = −G(t). Now G is “odd after a shift.”
Examples of the “odd after a shift” idea
– f(x) = sin(x − a) has rotational symmetry about (a, 0), because sin is odd at 0:
f(2a − x) = sin(a − x) = −sin(x − a) = −f(x) = 2·0 − f(x).
– Any non-vertical line y = m(x − a) + b has rotational symmetry about every point on the line:
f(2a − x) = m((2a − x) − a) + b = m(a − x) + b = 2b − (m(x − a) + b) = 2b − f(x).
A couple of useful side notes
– Horizontal-line symmetry y = b for a function y = f(x) forces f(x) = b, i.e., the function must be constant. Otherwise a vertical graph can’t mirror across a horizontal line and still be the same graph of a single-valued function.
– Domain matters. For symmetry about x = a, the domain must be closed under x ↦ 2a − x.
Short summary
– Symmetric about x = a: f(x) = f(2a − x). Equivalently, f(a + t) = f(a − t).
– Rotationally symmetric about (a, b): f(2a − x) = 2b − f(x). Equivalently, f(a + t) − b is odd.
External explanation
– Purplemath’s page on even and odd functions gives a clear symmetry viewpoint, and you can apply the same logic after shifting by a: https://www.purplemath.com/modules/fcnparity.htm
– For shifting functions to recenter at x = a, see Khan Academy’s shifting-intro: https://www.khanacademy.org/math/algebra2/manipulating-functions/shifting-functions-intro/a/shifting-functions-intro
So your instinct “h(2 + t) equals h(2 − t)” is exactly the right mirror. Your brain was not flipping left and right; it was just standing on a moving walkway at x = a.
You’ve got it: to test symmetry about x = a, shift first and check evenness-f(a + t) = f(a − t) for all t (so h(2 + t) = t² = h(2 − t)).
For an “odd-ish” version, use f(a + t) = −f(a − t), and around a point (a, b) you’d check f(a + t) = b − f(a − t) for a 180°-spin vibe.
Yep-when the mirror is x = a, the test is f(2a − x) = f(x) (same as f(a + t) = f(a − t)); for h(x) = (x − 2)^2, you get h(4 − x) = h(x), like folding the graph on a paper crease at x = 2. For an “odd around a point” vibe, check f(a + t) − b = −(f(a − t) − b) (and some folks loosely say “odd about the line” as f(2a − x) = −f(x), which is like folding at x = a and flipping the page).