# Inverse function of a parabola

1. Nov 15, 2007

### fatou123

from a graph of a function ( i obtained the graphg by doing a translation and y-scaling) g(x)= 1/3(x-2)^2 -3 b (x in [2:5]) i can see that g is increasing and so it is a one-one function and an image set is [-3;0]. so therefore the function g has an unverse function g^-1 .

so i can find the rule of g^-1 by solving this equations:

y=g(x)=1/3 (x-2)^2 -3

to obtain x in term of y

i have y=1/3(x-2)^2 -3 that is x= +or-3sqrt-1/9 - 1/3y is this right ?

Last edited: Nov 15, 2007
2. Nov 15, 2007

### HallsofIvy

Staff Emeritus
No, the whole point of being "one-to-one" is that you have one value, not two.
Unfortunately, you didn't show HOW you solve for x so I can't comment but that surely does not look right! (It's hard to be sure since you don't use parentheses to show what you really mean.) You are correct that when x= 2, y= -3. If you take y= -3 in the formula you give, do you get x= 0? When you take x= 5, y= g(5)= 0. When you take y= 0 in your equation, do you get x= 5?

If y= (1/3)(x-2)^2- 3 with x between 2 and 5, then y+ 3= (1/3)(x-2)^2, (x-2)^2= 3(y+3) x-2= sqrt(3(y+3)) and finally x= 2+ sqrt(3(y+3)). The PLUS is used rather than the MINUS because x must be larger than 2. Finally, don't forget to write the solution itself:
g-1(x)= 2+ sqrt(3(x+3)).

3. Nov 16, 2007

### fatou123

thank hallsofivy you have just confirm what i though that theyr were something wrong in my result =.

i tackle this equation by comoleting thr squares of a quadratic eqautions,and i think this is where i got confused by the two result i was looking for! lol
the domain of g(x) is given to us as x in [2;5] so i can't really argue this.

i solved the g^-1 by rearranging g(x)

y=1/3(x-2)^2 -3 , 1/3(x-2)^2=-3-y , (x-2)^2= -1/9-1/3y, x-3= SQRT-1/9-1/3y hence
x=3 SQRT-1/3-1/3y

thank you again i undersrtood were i got wrong in this it was quite obvious really lol

4. Nov 16, 2007

### HallsofIvy

Staff Emeritus
This first step is incorrect.