Finding the Equation of a Reflected GraphDate: 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/ |
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