Finding where point (m, n) ends up when there's an inverse transformation.

[AgeOfAsparagus]

My son is taking Pre-Calculus 12 (in BC), and we need help with this question:

If the point (m, n) is on the curve y = f(x) where does (m, n) end up after these transformations?

[math]y = \frac{1}{2}f^{-1} (x+5)+8[/math]
Here is what we tried, which is wrong:

[math]m \rightarrow n-5[/math],
[math]n \rightarrow \frac{1}{2}m + 8[/math]
The normal adjustments, but swapping m and n to account for the inversion.

I can't find any examples of this kind of question online (e.g. Khan academy). Lots of absolute value examples, and it should be the same process, but I can't figure it out.

Thank you!

 

[blamocur]

There might be some convention/language that I am not familiar with, but the above question does not make sense to me. The transformation is from 1D to 1D, but the point [imath](m,n)[/imath] is 2D. I.e. how can the transformation be applied to a 2D point?

 

[Dr.Peterson]

My son is taking Pre-Calculus 12 (in BC), and we need help with this question:

If the point (m, n) is on the curve y = f(x) where does (m, n) end up after these transformations?

[math]y = \frac{1}{2}f^{-1} (x+5)+8[/math]
Here is what we tried, which is wrong:

[math]m \rightarrow n-5[/math],
[math]n \rightarrow \frac{1}{2}m + 8[/math]
The normal adjustments, but swapping m and n to account for the inversion.

I can't find any examples of this kind of question online (e.g. Khan academy). Lots of absolute value examples, and it should be the same process, but I can't figure it out.

Thank you!

I don't like the wording (because they gave you an equation, not a set of transformations); but as I understand it, they are saying that the graph of [imath]y = \frac{1}{2}f^{-1} (x+5)+8[/imath] is obtained by applying certain transformations to the graph of [imath]y=f(x)[/imath], which are implied by the form in which it is written. If (m, n) is a point on the graph of [imath]y=f(x)[/imath] (so that [imath]f(m) = n[/imath]), what point do those transformations take that point into?

I would start by observing that [imath]m = f^{-1}(n)[/imath]. What value of x lets you use this fact to find y? (We want [imath]x+5 = n[/imath].) What value of y do you get? (We get [imath]y=\frac{1}{2}m + 8[/imath].) That will be the point you want. I obtain what you got, namely the point [imath](n-5, \frac{1}{2}m + 8)[/imath].

Why do you say your answer is wrong? If a computer rejected your answer, did you make sure to enter it in the form required? What was it that you gave as the answer?

 

[AgeOfAsparagus]

I don't like the wording (because they gave you an equation, not a set of transformations); but as I understand it, they are saying that the graph of [imath]y = \frac{1}{2}f^{-1} (x+5)+8[/imath] is obtained by applying certain transformations to the graph of [imath]y=f(x)[/imath], which are implied by the form in which it is written. If (m, n) is a point on the graph of [imath]y=f(x)[/imath] (so that [imath]f(m) = n[/imath]), what point do those transformations take that point into?

I would start by observing that [imath]m = f^{-1}(n)[/imath]. What value of x lets you use this fact to find y? (We want [imath]x+5 = n[/imath].) What value of y do you get? (We get [imath]y=\frac{1}{2}m + 8[/imath].) That will be the point you want. I obtain what you got, namely the point [imath](n-5, \frac{1}{2}m + 8)[/imath].

Why do you say your answer is wrong? If a computer rejected your answer, did you make sure to enter it in the form required? What was it that you gave as the answer?

I double checked my son's homework and this is NOT the answer he put in his assignment. *FACEPALM* Thank you for the confirmation!