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Symmetry of a graph

  • #1

Homework Statement



How can I tell if the following equation is symmetrical and whether it is even or odd?

Homework Equations



f(x) = x^2 - 6x

The Attempt at a Solution



I would guess that it is symmetric about x = 6 but that's not what the book says.
 

Answers and Replies

  • #2
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Homework Statement



How can I tell if the following equation is symmetrical and whether it is even or odd?

Homework Equations



f(x) = x^2 - 6x
Other relevant information would include the definitions of "even" and "odd" functions. How are these terms defined?

The Attempt at a Solution



I would guess that it is symmetric about x = 6 but that's not what the book says.
Have you graphed this equation? The graph would show that this parabola is not symmetric about the line x = 6.
 
  • #3
epenguin
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Might be cheating but if you know how to find the minimum on the graph, as (I tell you) there is only one, IF if is symmetrical it would have to be symmetrical about that, do you see that?
 
  • #4
Other relevant information would include the definitions of "even" and "odd" functions. How are these terms defined?
The graph of an even function is symmetric about the y-axis. The graph of an odd function is symmetric about the origin.

Have you graphed this equation? The graph would show that this parabola is not symmetric about the line x = 6.
I'm thinking that I have to complete the square?
 
  • #5
eumyang
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The graph of an even function is symmetric about the y-axis. The graph of an odd function is symmetric about the origin.
I think Mark44 is asking for another definition, one that involves function notation (ie. f(x), or, to give you a hint, f(-x)). Look in your book and try again.


EDIT: I was right (see below).
 
  • #6
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There are definitions of these terms that don't involve the graphs.
A function is even if f(-x) = f(x) for all x in the domain of f.
A function is odd if f(-x) = -f(x) for all x in the domain of f.

Completing the square will give you information about the vertex of this parabola.
 
  • #7
Redbelly98
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I'm thinking that I have to complete the square?
Yes, and that will get the quadratic function in vertex form, which will be useful for answering the question about the symmetry.
 
  • #8
epenguin
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Or another way. f(x) = 0 , for your f is very easy to solve - your f has been factorised already for you. So you have got a value of f for which there are two points, two values of x. Call them x1 and x2 to be more general. So if f is symmetrical it has to be symmetrical for these two points. I.e. it has to be symmetrical around what x-value (call it xm)? That would be true for any function f. The converse is not true for any function in general but it is true in general that an axis of symmetry can't be any other value of x than xm the one found as above*, so you will narrow it down vastly by finding that. Once you have you can probably show that x = xm is a line of symmetry for your whole function in your case (sufficiency as well as necessity).

To know what I'm talking about it may be better to draw a graph of you function. Quite an important problem because it is the start of more general and extensive principles.

*i.e. the condition is necessary but not sufficient
 

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