Symmetry in Graphs
The two kinds of symmetry from the previous lesson apply unchanged to graphs. What is new is that a graph comes with an equation, so symmetry can be established by substitution rather than by looking, which turns a guess into a proof.
In $y = x^2$, replace $x$ by $-x$. Does the equation change? Now do the same in $y = x^3$. What is different, and what would that difference look like on the two graphs?
To test for line symmetry in the y-axis, replace $x$ by $-x$ and see whether the equation is unchanged. To test for rotational symmetry of order 2 about the origin, replace $x$ by $-x$ and $y$ by $-y$ and see whether the equation is unchanged.
$$x \to -x \quad \text{or} \quad (x, y) \to (-x, -y)$$
The tests work because a symmetry means the set of points is unchanged. Reflecting in the y-axis sends the point $(x, y)$ to $(-x, y)$, so the curve is symmetric exactly when $(-x, y)$ satisfies the equation whenever $(x, y)$ does.
Know
- that a graph has line symmetry in the y-axis when replacing $x$ by $-x$ leaves the equation unchanged
- that a graph has rotational symmetry of order $2$ about the origin when replacing both $x$ by $-x$ and $y$ by $-y$ leaves the equation unchanged
- the symmetry of the standard graphs: parabolas, cubics, the hyperbola $y = \dfrac{1}{x}$, circles and straight lines
Understand
- why the substitution test works, in terms of the image of a point under the transformation
- why a parabola with its vertex off the y-axis is still symmetric, about a different vertical line
- why a single counterexample disproves symmetry while examples alone cannot prove it
Can Do
- test a given equation for line symmetry in the y-axis and for rotational symmetry about the origin
- state the axis of symmetry of a parabola given in the form $y = (x - h)^2 + k$
- describe the symmetry of a circle centred at the origin and of the hyperbola $y = \dfrac{1}{x}$
The symmetry of a graph means exactly what it meant for a shape: some reflection or rotation maps the graph onto itself.
What is new is that a graph is described by an equation, so "maps onto itself" can be checked without drawing anything. A point lies on the graph exactly when its coordinates satisfy the equation, so the graph is unchanged by a transformation precisely when the transformed coordinates also satisfy it.
Reflection in the y-axis sends $(x, y)$ to $(-x, y)$. So the graph has that symmetry exactly when $(-x, y)$ satisfies the equation whenever $(x, y)$ does, which happens exactly when replacing $x$ by $-x$ leaves the equation unchanged.
The whole method is that one sentence, applied to different transformations.
Replace every $x$ with $-x$, simplify, and compare with the original.
For $y = x^2 - 4$: replacing gives $y = (-x)^2 - 4 = x^2 - 4$, which is the original. So the graph is symmetric in the y-axis.
For $y = x^3$: replacing gives $y = (-x)^3 = -x^3$, which is not the original. So it is not symmetric in the y-axis.
The reason even powers survive and odd powers do not is simple arithmetic: $(-x)^2 = x^2$ but $(-x)^3 = -x^3$. A polynomial made only of even powers of $x$, constants included, always has this symmetry, and one containing any odd power does not.
The absolute value graph $y = |x|$ has it too, for the same reason: $|-x| = |x|$.
A rotation of $180°$ about the origin sends $(x, y)$ to $(-x, -y)$. So replace $x$ by $-x$ and $y$ by $-y$, simplify, and compare.
For $y = x^3 - x$: the substitution gives $-y = (-x)^3 - (-x) = -x^3 + x$. Multiplying both sides by $-1$ gives $y = x^3 - x$, the original. So the graph has rotational symmetry of order $2$ about the origin.
For $y = x^2$: the substitution gives $-y = x^2$, that is $y = -x^2$, which is a different curve. So it does not.
Note the tidy step of multiplying through by $-1$. Reaching $-y = -x^3 + x$ and stopping would suggest failure; the equation is unchanged, but you have to finish rearranging it to see so.
A parabola always has line symmetry, but the axis is only the y-axis when the vertex sits on it.
The graph of $y = (x - 3)^2 + 1$ has its vertex at $(3, 1)$ and is symmetric about the vertical line $x = 3$. The substitution test for the y-axis correctly reports failure, because that is not where the symmetry is.
To confirm the real axis, compare the values at equal distances either side of it. Taking $x = 3 + t$ gives $y = t^2 + 1$, and taking $x = 3 - t$ gives $y = (-t)^2 + 1 = t^2 + 1$. Equal outputs for equal steps in each direction is exactly what symmetry about $x = 3$ means.
So a failed y-axis test does not mean no line symmetry. It means no line symmetry in the y-axis, which is a narrower claim, and the two are easily confused.
Worth knowing without testing each time.
$y = x^2$ and any even-powered polynomial: line symmetry in the y-axis, no rotational symmetry about the origin.
$y = x^3$, $y = x$ and any odd-powered polynomial with no constant term: rotational symmetry of order $2$ about the origin, no line symmetry in the y-axis.
$y = \dfrac{1}{x}$: rotational symmetry of order $2$ about the origin, and line symmetry in both $y = x$ and $y = -x$.
A circle centred at the origin: infinitely many axes, every line through the centre, and rotational symmetry of every order.
A straight line through the origin: rotational symmetry of order $2$, and line symmetry in itself and in the perpendicular through the origin.
A straight line not through the origin still has line symmetry in itself, which is a slightly odd case worth noticing: the axis of symmetry is the graph.
Watch Me Solve It · 3 examples
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1State what must be shownReflection in the y-axis sends $(x, y)$ to $(-x, y)$, so the equation must be unchanged when $x$ is replaced by $-x$.
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2Substitute$y = (-x)^2 - 4$Every occurrence of $x$ is replaced, and brackets are kept until the power is evaluated.
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3Simplify and compare$y = x^2 - 4$Since $(-x)^2 = x^2$, the result is identical to the original equation.
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4ConcludeThe equation is unchanged, so whenever $(x, y)$ is on the graph so is $(-x, y)$. The graph therefore has line symmetry in the y-axis.
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1Substitute for the rotation$-y = (-x)^3 - (-x)$A half turn about the origin sends $(x, y)$ to $(-x, -y)$, so both variables change sign.
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2Simplify the right-hand side$-y = -x^3 + x$Since $(-x)^3 = -x^3$ and subtracting a negative gives a plus.
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3Multiply through by minus one$y = x^3 - x$This is the original equation, so the graph is unchanged by the half turn and the order is $2$.
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4Test the y-axis separately$y = (-x)^3 - (-x) = -x^3 + x$This is not the original, so there is no line symmetry in the y-axis. A single point confirms it: $(2, 6)$ is on the graph but $(-2, 6)$ is not, since $(-2)^3 - (-2) = -6$.
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1Test the y-axis first$y = (-x - 3)^2 + 1 = (x + 3)^2 + 1$This is not the original equation, so there is no line symmetry in the y-axis.
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2Locate the vertexThe squared term is zero when $x = 3$, and the square is never negative, so the smallest value of $y$ is $1$ and it occurs at $x = 3$. The vertex is $(3, 1)$.
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3Test the vertical line through the vertex$x = 3 + t: \quad y = t^2 + 1$$x = 3 - t: \quad y = (-t)^2 + 1 = t^2 + 1$Equal distances either side of $x = 3$ give equal outputs, which is what symmetry about that line means.
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4Check for rotational symmetryA parabola has no rotational symmetry beyond the full turn, because a half turn would send its single minimum to a maximum, and the curve has no maximum. Its order is $1$.
Brain Trainer · 5 problems
Five items. State which symmetry, if any, each graph has about the axes or the origin.
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1 Does $y = x^4 + 2$ have line symmetry in the y-axis?
$(-x)^4 = x^4$, so the equation survives.Yes -
2 Does $y = x^5$ have rotational symmetry of order $2$ about the origin?
$-y = (-x)^5 = -x^5$, which rearranges to the original.Yes -
3 Does $y = x^2 + x$ have line symmetry in the y-axis?
The substitution gives $x^2 - x$, which is different. The odd power spoils it.No -
4 What is the axis of symmetry of $y = (x + 5)^2 - 2$?
The squared term vanishes at $x = -5$, which locates the vertex.$x = -5$ -
5 How many axes of symmetry does the circle $x^2 + y^2 = 25$ have?
Every line through the centre works, and the centre is the origin.Infinitely many
Multiple Choice · 5 questions
To test whether a graph has line symmetry in the y-axis, you should replace:
Which graph has line symmetry in the y-axis?
The graph of $y = \dfrac{1}{x}$ has:
The axis of symmetry of $y = (x - 4)^2 + 3$ is:
A graph fails the test for line symmetry in the y-axis. It follows that the graph:
Short Answer · 3 questions
Substitute the image of a point
A symmetry maps the graph onto itself, so substituting the transformed coordinates must leave the equation unchanged. That single principle generates every test.
The two standard tests
Replace $x$ by $-x$ for line symmetry in the y-axis. Replace $x$ by $-x$ and $y$ by $-y$ for rotational symmetry of order $2$ about the origin.
Even and odd powers decide it
Only even powers of $x$, plus constants, gives y-axis symmetry. Only odd powers, with no constant, gives the half-turn symmetry about the origin.
A failed test is a narrow conclusion
Failing the y-axis test rules out that line only. A parabola with its vertex elsewhere is still symmetric, about the vertical line through its vertex.
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