The distance between the lines  and  is 6 units. Therefore, the first reflection should be 6 units below .  Therefore, the first reflection is:

The distance between  and  is 7 units.  Therefore, second reflection should be 7 units above .

The result is .

The result of reflecting  across  will become .  Reflecting the equation  across  will become .

The correct answer is .

First, if  was reflected across , the new line will be .

Reflecting this line  across  will yield .

Reflecting  across  will yield .

The answer is .

So far all of the transformations have come after or inside the function. But what about those that come before the function? This is the basis of reflections which can be made to reflect across the x or the y axis. Since a function takes an x value and returns a y value, the placement of our sign matters. For instance,  is a parabola opening up whose vertex is at the origin. But if one was to place a  -  sign in front of the fucntion like this , then the function would always return a negative y value. Because of this, a negative sign in front of the function, but not "in" it will reflect over the x axis. This is equivalent to turning the first function,,  a smile, into a frown. Use a graphing utility to visualize this. Now what if you placed a negative sign within the function? This would automatically make all x values negative. And since a function returns a y value and we will always be starting from the negative side of the x axis, this would reflect across the y axis. Use the function  to visualize this in a graphing utility.

 is the correct answer because a negative 1 inside the function (which is within the square root) shifts the function to the right (it is opposite of intuition) one unit and outside the function a negative 3 shifts the function 3 units down. Finally, because our negative sign is outside the function we will always yield negative y values and have thus reflected over the x axis. 

One way to determine algebraically if a function is an even function, or symmetric about the y-axis, is to substitute  in for . When we do this, if the function is equivalent to the original, then the function is an even function. If not, it is not an even function. 

For our function: 

Thus the function is not symmetric about the y-axis. 

One way to determine algebraically if a function is an even function, or symmetric about the y-axis, is to substitute  in for . When we do this, if the function is equivalent to the original, then the function is an even function. If not, it is not an even function. 

For our function: 

Since this matches the original, our function is symmetric across the y-axis. 

In order to determine if there is symmetry about the x-axis, replace all  variables with .   Solving for , if the new equation is the same as the original equation, then there is symmetry with the x-axis.

Since the original and new equations are not equivalent, there is no symmetry with the x-axis.

The correct answer is:

For a function to be even, it must satisfy the equality 

Likewise if a function is even, it is symmetrical about the y-axis 

Therefore, the function is not even, and so the answer is No

For a function to be symmetrical about the y-axis, it must satisfy   so there is not symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy   so there is symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x) and the resulting function must be equal to the original function.

-So there is no symmetry about the origin, and the answer is Symmetrical about the x-axis

For a function to be symmetrical about the y-axis, it must satisfy  so there is symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy  

 so there is not symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x) and the resulting function must be equal to the original function.

So there is no symmetry about the origin.