How Do You Convert a Transformed Function into Standard Form?

[aisha]

Rewrite Function PLZ HELP

I have a question for translation of functions
it says the equation of the image is -2 \sqrt(3x-12) -5
(note: sorry the square root should be over all of the 3x and -12 only -5 is outside of square root)
1) the first questions said what is the base function? I wrote \sqrt(x)

2) the second question said describe the series of transformations, so i wrote
Vertical stretch by a factor of 2, horizontal compression by a factor of 1/3, reflection in the x-axis, vertical translation 5 units down, horizontal translation 4 units left.

3) It says write the function into y=af[k(x-p)]+q form
I got -2\sqrt(3(x-(-4)) -5 (note: again square root is over everything in brackets except -5 )but i think this is wrong since there is a square root in my answer and no square root in the form that they want. Can someone help me out please Is only number 3 wrong? How do I rewrite the equation in the correct form?

 

[Diane_]

The only problem I can see with your #3 is that it should be (x-4), not (x-(-4)).

I only see one way to get rid of the square root, and that's more of a trick than anything else, but given the rest of your question it just may be what you need.

Let f(x) = your base function, \sqrt{x}. Then your function would be

y = 2\sqrt{3} f(x - 4) - 5

The square root is still there, it's just hidden in the "f(x)".

Does that look possible?

 

[aisha]

I don't understand

 

[Diane_]

All I've done is replace the \sqrt{x} with an f(x). If you replace the functional notation in my expression with \sqrt{x-4} and carry through the algebra, you should end up with your original expression.

I'm not entirely certain that's what you're looking for, but it's the only way I can see to make the square root go away. It seems like you have access to the answer - can you post that? It may make it easier to see exactly what you need to do.

 

[dextercioby]

I have a question for translation of functions
it says the equation of the image is -2 \sqrt(3x-12) -5
(note: sorry the square root should be over all of the 3x and -12 only -5 is outside of square root)
1) the first questions said what is the base function? I wrote \sqrt(x)

2) the second question said describe the series of transformations, so i wrote
Vertical stretch by a factor of 2, horizontal compression by a factor of 1/3, reflection in the x-axis, vertical translation 5 units down, horizontal translation 4 units left.

3) It says write the function into y=af[k(x-p)]+q form
I got -2\sqrt(3(x-(-4)) -5 (note: again square root is over everything in brackets except -5 )but i think this is wrong since there is a square root in my answer and no square root in the form that they want. Can someone help me out please Is only number 3 wrong? How do I rewrite the equation in the correct form?

How about writing it like that
y(x)=-2\sqrt{3x-12}-5=-2\sqrt{3(x-4)}-5=-2\sqrt{3}\sqrt{x-4}-5

which can be put under the form
y(x)=Af(x-4)-5
where
A=-2\sqrt{3}
f(x)=\sqrt{x}

What do you say now...??

Daniel.