Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Finding the Equation of a Reflected Graph

Date: 04/24/2004 at 14:51:02
From: Amanda and Bonny
Subject: When reflecting an equation, how does the equation change

When reflecting an equation in geometry, how does the equation change?
 For example, we were given the equation x squared plus y squared
quals nine.  We were told to find the change in the equation if the 
following transformation occured: a reflection over the x axis, and
then a reflection over the mirror of reflection y = 2.  How do we do
this without graphing the entire problem?

We understand how to graph the reflection, yet we are unsure how to 
find the way in which the equation changes without graphing the 
equation.  We know that when you reflect an equation over the x axis, 
all of the y values become the opposite of what they originally were.



Date: 04/24/2004 at 18:14:47
From: Doctor Greenie
Subject: Re: When reflecting an equation, how does the equation change

Hello, Amanda and Bonnie -

Let me first introduce some notation:  We usually use the "^" symbol 
to indicate exponentiation, so we write "x squared" as x^2 or "y to 
the fifth power" as y^5.  So in this problem our given equation is

  x^2 + y^2 = 9

I assume you know that this is a circle with center at the origin and 
radius 3.  And I understand that you know how to graph the reflections 
and that you want to know how to find the equations of the reflected 
graphs without doing the graphing--but you should do the graphing to 
make sure the answers you come out with for the reflected graphs are 
reasonable.

One more item regarding notation: it is common to use the "'" symbol 
to represent the image of an x- or y-coordinate after a 
transformation--so we can denote the image of (x,y) under a 
transformation as (x', y').

When we reflect a graph about the x-axis, we know that, as you say,

  "the y values become the opposite of what they originally were"

And of course, the x-values do not change when we reflect about the 
x-axis.  So in a reflection about the x-axis, we have

  the x-coordinate stays the same:  x' = x
  the y-coordinate is replaced by its opposite:  y' = -y

So the image of a particular point (x,y) is the point (x,-y).

And to find the equation of the graph which is the reflection about 
the x-axis of a given equation, we simply replace "y" with "-y" in 
the given equation.  So with our example, we have, for the equation of 
the reflected graph (under the first reflection, about the x-axis)

  x^2 + (-y)^2 = 9

But (-y)^2 is the same as y^2 for all values of y, so this reflected 
equation is simply

  x^2 + y^2 = 9

This is the same as the original equation.  And we can see graphically 
that this should be the case--when we reflect a circle with its center
at the origin about the x-axis, the result is the same circle, because
of the symmetry of a circle.

Before we look at the second reflection, about the line y = 2, let's 
look at the reflection about the x-axis in a different way.  When we 
reflect a graph about the x-axis and look at a particular point on the 
graph and its reflection, we have been thinking of this as the y-value 
of the reflection being the opposite of the y-value of the original 
point.  Let's instead think of the original point and the reflected 
point being the same distance from the the line of reflection (in this 
case, the x-axis).  With those two points being the same distance from 
the x-axis, we can think of the value of y on the x-axis as being the 
average of the y-values of the original and reflected points.  Using 
this approach, we have, since y = 0 everywhere on the x-axis,

  (y + y')/2 = 0  [the average of the original and reflected
                    y-values is the y-value of the line of
                    reflection]

which leads us to the already familiar

  y' = -y

We can use this approach to understand what happens when we perform 
the second reflection, about the line y = 2.  Graphically, we can see 
that the center of the original circle, (0,0), has the point (0,4) as 
its reflection.  So our answer should give us an equation for a circle 
with center at (0,4) and radius 3.  Let's see what happens.

Using the approach just described, in a reflection about the line 
y = 2, the value y = 2 is the average of the y-values of any original 
point and its reflection; so we have

  (y + y')/2 = 2  [the average of the original and reflected
                    y-values is the y-value of the line of
                    reflection]

  y + y' = 4

  y' = 4 - y

And so, to find the equation of the graph after the second reflection, 
we simply replace "y" with "4 - y" in the original equation, to get

  x^2 + (4 - y)^2 = 9

Since (4 - y)^2 and (y - 4)^2 are the same, we can rewrite this 
equation in the more familiar form

  x^2 + (y - 4)^2 = 9

And this we do indeed recognize as the equation of a circle with 
center at (0,4) and radius 3--as we knew it should be.

I hope all this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Coordinate Plane Geometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/