Inverse functions and graphing: Find the inverse function for the function y = x^2 - 4x + 5, and draw its graph

[reggiwilliams]

Have been revising functions and came across this question
Made answer to part one (x-2)^2+1 and turning points (2,1)
Made answer to part two sqrt of (x-1) + 2
Are we agreed
However wasn’t sure how to graph inverse function given when x 0 upwards you have a square root of a negative number + 2 continuously
Can you help

 

[Dr.Peterson]

Have been revising functions and came across this question
Made answer to part one (x-2)^2+1 and turning points (2,1)
Made answer to part two sqrt of (x-1) + 2
Are we agreed
However wasn’t sure how to graph inverse function given when x 0 upwards you have a square root of a negative number + 2 continuously
Can you help

The trouble is that the "inverse" of the given function is not a function; properly it should be $\pm\sqrt{x-1}+2$, with two values; or else you could (arbitrarily) choose to restrict the domain of the function to make it one-to-one. Have you been taught a specific answer to give for such a question, or did you just forget the plus-or-minus? I suspect the intent of the problem is to call your attention to this issue.

I'm not sure what you mean by "when x 0 upwards you have a square root of a negative number + 2 continuously", but you probably mean something related to that. Can you show your graph, and explain your thinking in more detail? (Showing an image of your work may also help overcome your difficulty in typing the notation.)

 

[skeeter]

The inverse function will have its domain restricted to values of $x \ge 1$.

Why?

 

[reggiwilliams]

Forgot the - option which probably explains my query when entering a value for x of zero or less
Understand why inverse is not a function and therefore presumably cannot be drawn
Here are my workings and initial rough graph

 

[reggiwilliams]

The trouble is that the "inverse" of the given function is not a function; properly it should be $\pm\sqrt{x-1}+2$, with two values; or else you could (arbitrarily) choose to restrict the domain of the function to make it one-to-one. Have you been taught a specific answer to give for such a question, or did you just forget the plus-or-minus? I suspect the intent of the problem is to call your attention to this issue.

I'm not sure what you mean by "when x 0 upwards you have a square root of a negative number + 2 continuously", but you probably mean something related to that. Can you show your graph, and explain your thinking in more detail? (Showing an image of your work may also help overcome your difficulty in typing the notation.)

Forgot the minus answer
Presumably the inverse can’t be drawn
Here are my initial workings

 

[Dr.Peterson]

Forgot the minus answer
Presumably the inverse can’t be drawn
Here are my initial workings

Everything I see looks correct, if you have been taught to always choose the positive root for some reason. You have correctly graphed the function and the inverse of that function when restricted to x>=2 (that is, the inverse function restricted to y>=2). If you restricted the function to x<=2, the inverse would be $-\sqrt{x-1}+2$.

It's not that the inverse can't be drawn, but that it isn't fully defined; there are two possible inverse functions, and you chose one of them. (You could also draw the entire graph of $\pm\sqrt{x-1}+2$, but that isn't an inverse function. Its graph would be an entire horizontal parabola.)

​

The original function is red; your inverse is blue, and the negative inverse is green. The blue inverse is actually the inverse of this restricted function:

​

What is uncertain is what they actually expect of you, which may depend on knowing what you have been taught. If I were helping you in person, I would be looking in your textbook or notes to see what you have learned about this sort of situation. As I said, it is possible that they want you to wonder why your "inverse" is not an entire (reflected) parabola, and they will tell you in the next lesson; I just don't know.

Has the term "restricted domain" been introduced yet? Have you learned that the graph of the inverse is a reflection of the graph of the original function?

Forgot the - option which probably explains my query when entering a value for x of zero or less

Values of less than 1 simply are not in the domain of the inverse; that's not related to the plus or minus in your inverse. The domain of the inverse will be the range of the original function, which is [1,infinity).