Stuck on a few Transformations questions.

[Zxyxz]

Hi, I'm stuck on a few Transformation questions I have, I tried answering them but feel like i'm doing it wrong.


These are the questions ;

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These are the answers I wrote for the above;
a)y=1/3(3x-9)
b)y=3(4x-6)-2
c)y=3(1/6x+4)+5

For these questions

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The answers I wrote are;
a)y=√1/2(16-x² )
b)-y=√(3)(16-x²)


For this one I have no idea, but I still tried.

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Answer I wrote;
f-1(x)=-2+x/-5x


(Sorry I don't know how to format answers to make the proper signs.)

Any help on how to do these questions or what I'm doing wrong would be awesome, thanks!

 

[Ishuda]

What you have to learn is what is replace by what when you do a transformation in the context here. Looking at the first problem you seem to have the vertical down, that is if it is a factor of A for a vertical compression (0 < A < 1) or expansion (1 < A), you multiply the equation for y by A. That is, given
y = 3 x - 6
and you want to apply a compression of 1/3, you would have
y = (1/3) (3 x - 6) = x - 2
What you missed is the translation to the right. Say you are looking at something on a graph at x = 0 (in our example then, y would be -2). If you want to move it to the right you would want y to be minus 2 at say x = 3. If I replace x in the equation y = x-2 by x-3, and we have
y = (x-3) - 2 = x - 5
then when x is 3, y is -2 which is what we wanted.

So, you can remember what you need to do or work it out. For tests, remembering what to do is generally faster.
http://blogs.spsk12.net/4791/files/2012/10/2-7_2-6_2-5Notes.pdf
and
http://www.regentsprep.org/regents/math/algtrig/atp9/funclesson1.htm

 

[Zxyxz]

What you have to learn is what is replace by what when you do a transformation in the context here. Looking at the first problem you seem to have the vertical down, that is if it is a factor of A for a vertical compression (0 < A < 1) or expansion (1 < A), you multiply the equation for y by A. That is, given
y = 3 x - 6
and you want to apply a compression of 1/3, you would have
y = (1/3) (3 x - 6) = x - 2
What you missed is the translation to the right. Say you are looking at something on a graph at x = 0 (in our example then, y would be -2). If you want to move it to the right you would want y to be minus 2 at say x = 3. If I replace x in the equation y = x-2 by x-3, and we have
y = (x-3) - 2 = x - 5
then when x is 3, y is -2 which is what we wanted.

So, you can remember what you need to do or work it out. For tests, remembering what to do is generally faster.
http://blogs.spsk12.net/4791/files/2012/10/2-7_2-6_2-5Notes.pdf
and
http://www.regentsprep.org/regents/math/algtrig/atp9/funclesson1.htm

Thanks, that made a lot of sense. Just wondering but for the first answer, y=(x-3)-2 = x-5, can't I just add the x-3 and x-5?

for b) I first did the compression to make it y=(7x-6), reflected it by the Y axis and got y=(-7x-6), and finally 2 units down would make it y=(-7x-6)-2.
for c) I did the expansion by 6 to make it y=(3)(x/6-6), 4 units up would be y=(3)(x/6-6)=x+4, and finally 5 units up which made it y=(3)(x/6-6)=(x+4)+5.

I still don't understand how I'd do the other 2 questions, does the square root play a role in determining the equation for the new graph?

And the last one I have no idea how to inverse, I just assume based on what the teacher taught that everyone would be reversed.