Solving backwards for invariant point of f(x) = x^2 - 6x - 18 under x = f(x)

[GuavaEater]

Hey there, this transformations unit has proven quite tough for me. The question I'm currently stuck on is "A point of f(x)=x²-6x-18 that is invariant for the transformation x=f(y) has an x value of:
A. -9
B. 0
C. 9
D. 2
Attached is the work I've done, and the original question. I'm really at a loss for where to even start, as the question itself confuses me. From what I understand, I'm solving for the invariant point, which would be likely more easy via graphing calculator. What I tried was inputting the original point function y₁ ( f(x)=x²-6x-18) ), as well as y₂ = x, but that hasn't yielded any obvious results. If anyone could get me started, or even explain more clearly what it's trying to ask, that'd be awesome.

 

[Harry_the_cat]

Hey there, this transformations unit has proven quite tough for me. The question I'm currently stuck on is "A point of f(x)=x²-6x-18 that is invariant for the transformation x=f(y) has an x value of:
A. -9
B. 0
C. 9
D. 2
Attached is the work I've done, and the original question. I'm really at a loss for where to even start, as the question itself confuses me. From what I understand, I'm solving for the invariant point, which would be likely more easy via graphing calculator. What I tried was inputting the original point function y₁ ( f(x)=x²-6x-18) ), as well as y₂ = x, but that hasn't yielded any obvious results. If anyone could get me started, or even explain more clearly what it's trying to ask, that'd be awesome.

One way to do this is to think about the graphs of y=f(x) and x=f(y) graphically.

You know for example that y=f(x)+2 moves the graph of y=f(x) up 2units...right?

You also know that y=-f(x) reflects the graph of y=f(x) in the x-axis...right?

Do you know what x=f(y) does to the graph of y=f(x)?

 

[GuavaEater]

I believe that means it's the inverse of the function. However, I'm not sure where to go from there.

 

[Harry_the_cat]

I believe that means it's the inverse of the function. However, I'm not sure where to go from there.

Yes it is the inverse function. What happens on a graph? What is the graphical relationship between a function and it's inverse?

 

[GuavaEater]

The co-ordinates of x and y are switched for each plot point. I'm just not sure how to apply that to the question, though. Sorry I'm so dense

 

[Harry_the_cat]

The co-ordinates of x and y are switched for each plot point. I'm just not sure how to apply that to the question, though. Sorry I'm so dense

No no. You're not dense. You're just in the process of learning! And that's a good thing!

Yes that's right, the x and y co-ordinates are swapped eg (2,7) becomes (7, 2) etc.

What sort of points wouldn't change then ie which points are invariant ie which points will stay the same after you swap the x and y coordinates?

(I'm making you work for this aren't I?)

 

[GuavaEater]

The points that have the same x and y co-ordinates, like (0,0), or (-1,-1). Those stay the same. For this graph, the invariant points are (-2,-2), and (9,9). So the answer would be 9?! I think I solved it, graphically!

 

[Harry_the_cat]

The points that have the same x and y co-ordinates, like (0,0), or (-1,-1). Those stay the same. For this graph, the invariant points are (-2,-2), and (9,9). So the answer would be 9?! I think I solved it, graphically!

Good. Any points on the line y=x (ie points where the y coord= xcoord) will be invariant for an inverse function. The graph of the inverse function is the original function reflected in the line y=x.

Algebraically you need to solve \(\displaystyle y=x^2-6x-18\) and \(\displaystyle y=x\) simultaneously to find which points on the original function line on the line of reflection( and hence are invariant).

Solving simultaneously:

\(\displaystyle y=x^2-6x-18\) .......1
\(\displaystyle y=x\)..........................2

Sub eqtn 1 into 2:

\(\displaystyle x^2-6x-18=x\)

\(\displaystyle x^2-7x-18=0\)

\(\displaystyle (x-9)(x+2)=0\)

\(\displaystyle x=9 \) or \(\displaystyle x=-2\)