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Chapter 1: Problem 19

Determine whether the graph of each equation is symmetric with respect to the \(y\) -axis, the \(x\) -axis, the origin, more than one of these, or none of these. $$x=y^{2}+6$$

Short Answer

Expert verified
The graph of the equation \(x=y^{2}+6\) is only symmetric with respect to the x-axis.

Step by step solution

01

Determine the symmetry with respect to the y-axis

The equation will be symmetric with respect to the y-axis if replacing x with -x gives the same equation. Let's apply this rule to \(x=y^{2}+6\). If we replace x with -x, it gives \(-x=y^{2}+6\), which is not the same as the original equation. Therefore, the equation is not symmetric with respect to y-axis.
02

Determine the symmetry with respect to the x-axis

The equation will be symmetric with respect to x-axis if replacing y with -y results in an identical equation. Let's apply this rule to \(x=y^{2}+6\). If we replace y with -y, it gives \(x=(-y)^{2}+6\), which simplifies to \(x=y^{2}+6\), the same as the original equation. So the equation is symmetric with respect to the x-axis.
03

Determine the symmetry with respect to the origin

An equation is symmetric with respect to the origin if replacing both x and y with their opposites results in the same equation. Let's apply this rule to \(x=y^{2}+6\). If we replace both x and y with -x and -y, it gives \(-x=(-y)^{2}+6\), which simplifies to \(-x=y^{2}+6\), not the same as the original equation. Therefore, the equation is not symmetric with respect to the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Y-Axis Symmetry
In mathematics, understanding graph symmetry helps in analyzing functions or relations visually. Y-axis symmetry, a common type, means that a graph mirrors itself across the y-axis. To check for this symmetry, substitute \(x\) with \(-x\) in the equation. If the equation remains unchanged, the graph is symmetric about the y-axis.

Let's consider the equation \(x = y^2 + 6\). By replacing \(x\) with \(-x\), we get \(-x = y^2 + 6\). This equation is not the same as the original, indicating that there is no y-axis symmetry. Remember:
  • If replacing \(x\) with \(-x\) results in the original equation, the graph has y-axis symmetry.
  • The reflection across the y-axis implies identical halves on either side.
Recognizing y-axis symmetry can simplify graphing and understanding the behavior of certain functions.
X-Axis Symmetry
X-axis symmetry occurs when a graph is reflected across the x-axis and remains identical. We test this by replacing \(y\) with \(-y\) in the equation. If the resulting equation matches the original, then x-axis symmetry is present.

Applying this to our equation \(x = y^2 + 6\), we substitute \(y\) with \(-y\), leading to \(x = (-y)^2 + 6\). This simplifies back to \(x = y^2 + 6\), confirming symmetry about the x-axis. Key points include:
  • If substituting \(y\) with \(-y\) gives the original equation, x-axis symmetry is confirmed.
  • This reflects the upper half of the graph onto the lower half across the x-axis.
Such symmetry helps in predicting graph behavior, making the equation easier to visualize and solve.
Origin Symmetry
Origin symmetry is a deeper concept in graph analysis, where the graph's reflection across both axes meets back on itself. To check for this, replace both \(x\) and \(y\) with their negative counterparts, \(-x\) and \(-y\). If the equation remains unchanged, origin symmetry is confirmed.

For \(x = y^2 + 6\), substituting both leads to \(-x = (-y)^2 + 6\). Simplifying gives \(-x = y^2 + 6\), which is not the original equation. Thus, this graph does not possess origin symmetry. Important aspects include:
  • If both \(x\) and \(y\) are replaced with negatives and the equation remains unchanged, origin symmetry exists.
  • Origin symmetry implies a 180-degree rotational similarity around the graph's center.
This understanding can be crucial for functions where rotational symmetry reveals important graph traits.

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Most popular questions from this chapter

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+x+y-\frac{1}{2}=0$$

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