By convention, we usually denote our 'pre-image' cartesian variables (that is, before transformation) as
)
.
Similarly, we usually denote our 'image' cartesian variables (that is, after transformation) as
)
For that particular question, we are transforming
^{\frac{1}{2}}+3\text{ to }y=-x^{\frac{1}{2}}+6)
As far as the notation described is concerned, that means we are transforming the pre-image
^{\frac{1}{2}}+3)
to the image
^{\frac{1}{2}}+6)
.
Transformations can be described by investigating how each individual cartesian variable is transformed. To do this, we need to compare each pre-image variable to its corresponding image variable. To do so, in each equation, we need to transpose to collect all of the vertical transformations on the same side of each equation as the vertical variable (and ensure all of the horizontal transformations are on the same side of the equation of the horizontal variable):
^{\frac{1}{2}}+3\rightarrow y'=-\left(x'\right)^{\frac{1}{2}}+6)
For the pre-image equation, we can subtract 3 from both sides, then divide both by 2. For the image equation, we can subtract 6 from both sides then divide both sides by -1.
^{\frac{1}{2}}\rightarrow -\left(y'-6\right)=\left(x'\right)^{\frac{1}{2}})
Now we equate each pre-image variable expression with its corresponding image variable expression to determine the transformations that have been performed on each variable (note that, from this point onwards, the function itself is irrelevant - we are doing this for a square root, but we can apply exactly the same principal to any other function in the course):
\text{ and }4-x=x')
To identify the transformations that have occurred, we need to find how we obtain each image variable from its pre-image counterpart. That is, we need to transpose each equation to get the image variable in terms of the pre-image variable:
Focusing on the vertical transformations:
Multiplying y by 1/2 dilates vertically (i.e. from the x-axis) by a factor of 1/2.
Multiplying y by a negative changes its sign (i.e. reflects in the x-axis).
Adding 15/2 to y translates all points 15/2 units up.
Now focusing on the horizontal transformations:
Multiplying x by a negative changes its sign (i.e. reflects in the y-axis).
Adding 4 to x translates all points 4 units right.
You can apply this technique to any "find the sequence of transformations..." type question. That is, for each of the following:
Find a sequence of transformations that takes the graph of y = -3(3x+1)^2 + 7 to the graph of y = x^2
Find a sequence of transformations that takes the graph of y = 2*(4-x)^(1/2) to (x)^(1/2)
Find a sequence of transformations that takes the graph of y = 2*(4-x)^(1/2)+3 to -(x)^(1/2)+6
We can use the same idea to apply transformations as well. For instance, for
y = 3/x^2
find the rule after the graph has undergone the following
1. dilation of factor 2 from the x-axis
2. Translation of 2 units in the negative direction of x-axis and 1 unit in the negative direction of the y-axis
3. Reflection in the x-axis
We have a pre-image equation of
and an unknown image equation.
We have transformation equations:
)
Note that in this case, because the reflection has been done last, I've applied it last (i.e. multiplication of all of 2y-1 by -1).
Because we want the image equation, we want our equation in terms of the image variables. If we transpose our transformation equations to get each pre-image variable in terms of their corresponding image variables, we can then substitute into the pre-image equation and transpose:
\\\Rightarrow x'+2=x\text{ and }\frac{-y'+1}{2}=y\\\therefore y=\frac{3}{x^2}\rightarrow\frac{-y'+1}{2}=\frac{3}{\left(x'+2\right)^2}\\\Rightarrow y'=-\left[2\left(\frac{3}{\left(x'+2\right)^2}\right)-1\right]\\\therefore y'=-\frac{6}{\left(x'+2\right)^2}+1)
So the answer that you obtained is indeed correct.
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