Determining the Line of Symmetry of a Reciprocal Equation.

  • Thread starter Samad
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Problem: How do you determine the line of symmetry of a reciprocal equation?

Solution:

For example, I'll graph the reciprocal function Y=1/(x+2)

eq.jpg

^Just a quick sketch

And the equation of the line of symmetry is simply -(x+2), which can be seen here:

eqsym.jpg

^Also a quick sketch

By adding a negative in front of any reciprocal equation, you have the line of symmetry. That much I figured out.

The Real Problem: Why does this happen? I mean, how can I explain it in plain english?
 

Answers and Replies

  • #2
HallsofIvy
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First, there is no such thing as "the" line of symmetry. Any hyperbola has two lines of symmetry. In the case you give, y= 1/(x-2), (Your graph is NOT of y= 1/(x+2). That would have a vertical asymptote at x= -2.), the center of the hyperbola is at (2, 0). The line you draw, through (2,0) and perpendicular to the line containing the vertices, so with slope -1, y= -(x-2)= 2- x, is one axis of symmetry. The line through (2,0) and passing through the vertices, y= x- 2, is also an axis of symmetry.

I am not at all sure what you mean by "By adding a negative in front of any reciprocal equation, you have the line of symmetry." Do you mean "if y= 1/(x-a) is a hyperbola, then y= -1(x-a)= a- x is a line of symmetry"? If that is what you mean, then, yes, that is true. For that particular kind of hyperbola, the two lines of symmetry have slopes 1 and -1 and pass through the center of the hyperbola, (a, 0). Their equations are, therefore, y= 1(x-a)= x-a and y=-1(x-a)= a- x.

More generally, any hyperbola has two lines of symmetry: the line through the two vertices and the line through the center (half way between the vertices) perpendicular to that line.
 

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