Problem 48
Question
45–50 ? Test the equation for symmetry. $$ x^{4} y^{4}+x^{2} y^{2}=1 $$
Step-by-Step Solution
Verified Answer
The equation is symmetric with respect to the x-axis, y-axis, and the origin.
1Step 1: Test for Symmetry with Respect to the x-axis
To test for x-axis symmetry, replace each occurrence of \( y \) in the equation with \( -y \). The equation becomes: \[ x^4 (-y)^4 + x^2 (-y)^2 = 1 \]. Since \((-y)^4 = y^4\) and \((-y)^2 = y^2\), the equation simplifies back to \[ x^4 y^4 + x^2 y^2 = 1 \]. The equation is symmetric with respect to the x-axis.
2Step 2: Test for Symmetry with Respect to the y-axis
To test for y-axis symmetry, substitute \( x \) with \( -x \) in the equation. This gives us: \[ (-x)^4 y^4 + (-x)^2 y^2 = 1 \]. Since \((-x)^4 = x^4\) and \((-x)^2 = x^2\), the equation remains: \[ x^4 y^4 + x^2 y^2 = 1 \]. The equation is symmetric with respect to the y-axis.
3Step 3: Test for Symmetry with Respect to the Origin
For origin symmetry, replace both \( x \) and \( y \) with \( -x \) and \( -y \) respectively. The equation becomes: \[ (-x)^4 (-y)^4 + (-x)^2 (-y)^2 = 1 \]. As previously calculated, \((-x)^4 = x^4\) and \((-y)^4 = y^4\); similarly, \((-x)^2 = x^2\) and \((-y)^2 = y^2\). The equation simplifies back to: \[ x^4 y^4 + x^2 y^2 = 1 \]. The equation is symmetric with respect to the origin.
4Step 4: Conclusion: Symmetry Analysis Summary
The given equation \( x^4 y^4 + x^2 y^2 = 1 \) is symmetric with respect to the x-axis, y-axis, and the origin.
Key Concepts
Symmetric Functionx-axis Symmetryy-axis SymmetryOrigin Symmetry
Symmetric Function
A symmetric function is a special type of function that maintains its form or expression unchanged under certain transformations. Often, these transformations involve algebraic operations such as reflections about axes or the origin in a coordinate plane.
In the context of the equation \(x^{4}y^{4} + x^{2}y^{2} = 1\), symmetric functions can be identified by testing the equation for symmetry properties.
In the context of the equation \(x^{4}y^{4} + x^{2}y^{2} = 1\), symmetric functions can be identified by testing the equation for symmetry properties.
- An equation is symmetric with respect to an axis if substituting the variables with their negatives leaves the equation unchanged.
- Origin symmetry involves more comprehensive transformations, testing the equation's behavior when both variables are negated simultaneously.
x-axis Symmetry
Testing for x-axis symmetry involves checking whether a function mirrors itself over the x-axis. Technically, this means if you substitute \(y\) with \(-y\) and the equation remains unchanged, it exhibits x-axis symmetry.
In our example equation \(x^4 y^4 + x^2 y^2 = 1\), when we replace \(y\) with \(-y\), we get \(x^4 (-y)^4 + x^2 (-y)^2 = 1\).
Applying properties of exponents, since both even powers of \(-y\) result simply in \(y^4\) and \(y^2\), the equation simplifies back to its original form.
In our example equation \(x^4 y^4 + x^2 y^2 = 1\), when we replace \(y\) with \(-y\), we get \(x^4 (-y)^4 + x^2 (-y)^2 = 1\).
Applying properties of exponents, since both even powers of \(-y\) result simply in \(y^4\) and \(y^2\), the equation simplifies back to its original form.
- This invariance shows that the equation’s graph is unchanged when flipped over the x-axis.
- It's a clear sign of x-axis symmetry, helping identify and understand functions visually.
y-axis Symmetry
Y-axis symmetry is identified when substituting \(x\) with \(-x\) keeps an equation unchanged. This signifies the equation's shape is mirrored across the y-axis.
The equation \(x^4 y^4 + x^2 y^2 = 1\) demonstrates y-axis symmetry when we substitute \(x\) with \(-x\).
The expression \((-x)^4 y^4 + (-x)^2 y^2 = 1\) simplifies back to the original equation due to the even powers, where \((-x)^4 = x^4\) and \((-x)^2 = x^2\).
The equation \(x^4 y^4 + x^2 y^2 = 1\) demonstrates y-axis symmetry when we substitute \(x\) with \(-x\).
The expression \((-x)^4 y^4 + (-x)^2 y^2 = 1\) simplifies back to the original equation due to the even powers, where \((-x)^4 = x^4\) and \((-x)^2 = x^2\).
- This test confirms that the left and right sides of the graph are reflections, enhancing its identification visually and algebraically.
- An understanding of y-axis symmetry assists in sketching and solving algebraic graphs effectively.
Origin Symmetry
Graphical symmetry concerning the origin can be tested by replacing each variable \(x\) and \(y\) with their negatives, such as \(-x\) and \(-y\), respectively. An equation maintaining its form under these transformations is said to have origin symmetry.
For the equation \(x^4 y^4 + x^2 y^2 = 1\), replacing \(x\) and \(y\) simultaneously with \(-x\) and \(-y\) results in \((-x)^4 (-y)^4 + (-x)^2 (-y)^2 = 1\).
Since even powers negate the influence of the negative sign, this simplifies back to its original form.
For the equation \(x^4 y^4 + x^2 y^2 = 1\), replacing \(x\) and \(y\) simultaneously with \(-x\) and \(-y\) results in \((-x)^4 (-y)^4 + (-x)^2 (-y)^2 = 1\).
Since even powers negate the influence of the negative sign, this simplifies back to its original form.
- Origin symmetry indicates that flipping the graph over both axes results in a graph unchanged in appearance.
- This property helps confirm that each quadrant reflects the pattern of others, simplifying analysis and sketching of the function.
Other exercises in this chapter
Problem 48
If \(M(6,8)\) is the midpoint of the line segment \(A B,\) and if \(A\) has coordinates \((2,3),\) find the coordinates of \(B .\)
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Find all real solutions of the equation, correct to two decimals. $$ x^{4}=16-x^{3} $$
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Find the slope and \(y\)-intercept of the line and draw its graph. \(3 x-4 y=12\)
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(a) Sketch the parallelogram with vertices \(A(-2,-1)\) \(B(4,2), C(7,7),\) and \(D(1,4) .\) (b) Find the midpoints of the diagonals of this parallelogram. (c)
View solution