# Simple Transformations Questions

Sorry I don't know how to show the root sign on the forums so I am just gonna use /

1. Homework Statement
A stretch is applied to the graph of y=/x to produce the graph of y=/9x . Relative to the x and y axis, this stretch may be described as either a........ (then it lists a, b, c, or d answers)

horizontal stretch by a factor of 1 over 9 about the y-axis or a vertical stretch by a factor of 3 about the x-axis.

2. The attempt at a solution
I understand that the root of 9 is 3, so it makes sense to me that the vertical stretch is 3. But what I don't understand is why is the horizontal stretch is 1 over 9. Should I not find the root of that and then it be 1 over 3? The only thing I can think of is the the 9 is like this /(9x) and I have to take the 9 out and then I still have a nine outside. /9 = 3 then 3^2 = 9 again. So it comes out to be 1 over 9.

I also assume, although I'm not sure if I'm correct is the reason that it can be a vertical or horizontal stretch is because its not defined in the equation as being a vertical or horizontal stretch.

Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
Sorry I don't know how to show the root sign on the forums so I am just gonna use /

1. Homework Statement
A stretch is applied to the graph of y=/x to produce the graph of y=/9x . Relative to the x and y axis, this stretch may be described as either a........ (then it lists a, b, c, or d answers)

horizontal stretch by a factor of 1 over 9 about the y-axis or a vertical stretch by a factor of 3 about the x-axis.

2. The attempt at a solution
I understand that the root of 9 is 3, so it makes sense to me that the vertical stretch is 3. But what I don't understand is why is the horizontal stretch is 1 over 9. Should I not find the root of that and then it be 1 over 3? The only thing I can think of is the the 9 is like this /(9x) and I have to take the 9 out and then I still have a nine outside. /9 = 3 then 3^2 = 9 again. So it comes out to be 1 over 9.

I also assume, although I'm not sure if I'm correct is the reason that it can be a vertical or horizontal stretch is because its not defined in the equation as being a vertical or horizontal stretch.
Obviously $y= \sqrt{9x}$ is the same thing as $y= 3\sqrt{x}$. Since your base equation is $y'= \sqrt{x'}$ (I put the primes on to distinguish the "old" equation from the new), thinking of this a "change in y" it is $y= 3y'= 3(\sqrt{x})$: the "old value of y" is multiplied by 3 so it is a "stretch" of 3 of the vertical y-axis. However, thinking of this as a change only in x, $y= \sqrt{x'}= \sqrt{9x}$: x'= 9x or x= (1/9)x'. The "old value" of x' is multiplied by 1/9, not 1/3. Changes in x occur before the "base" function (here $\sqrt{x}$) is applied, changes in y occur after.

(Click on the equations to see the Tex code.)

Ok thanks for clarifying that for me.