What are the Inverses of y=x^{2}+4x-6?

[crybllrd]

Homework Statement

The function

$y=x^{2}+4x-6$

has two inverses. What are they and which domains lead to these inverses?

Homework Equations

The Attempt at a Solution

$y=x^{2}+4x-6$

$x=y^{2}+4y-6$

$y(y+4)=x+6$

Not really sure where to go from here.

 

[Mark44]

Homework Statement

The function

$y=x^{2}+4x-6$

has two inverses. What are they and which domains lead to these inverses?

Homework Equations

The Attempt at a Solution

$y=x^{2}+4x-6$

Write this equation as x² + 4x - 6 - y = 0, and solve for x using the quadratic formula. That will give you x = f^-1(y) (with some abuse of notation as f^-1 is not a function).

$x=y^{2}+4y-6$

$y(y+4)=x+6$

This is no help at all.

Not really sure where to go from here.

 

[Mentallic]

The Attempt at a Solution

$y=x^{2}+4x-6$

$x=y^{2}+4y-6$

$y(y+4)=x+6$

Not really sure where to go from here.

That's similar to being asked to solve the quadratic

$$x^2+2x=3$$

and then taking the next step as follows, and getting stuck

$$x(x+2)=3$$

You need to factorize! Or if you can't, which will be the case if you have a number such as, say, 4 instead of the 3, then you need to use the quadratic formula. If the 3 is replaced with a constant or variable, such as

$$x^2+2x=k$$

then you definitely need to use the quadratic formula, applying all the same rules you know, but simply extending it to the realm outside of mere known constants.

 

[crybllrd]

OK, I got it from here I believe. I'm on mobile, so I will work it out when I get home and post my answer.

 

[crybllrd]

Alright, so I used the following for the quadratic formula:

$a=1, b=4, and, c=(-6-y)$

to get

$x=-2\pm2\sqrt{10+y}$

That will give you x = f$^{-1}$(y)

At this point, would I swap x and y to get the inverse of f(x)?

 

[Mentallic]

Alright, so I used the following for the quadratic formula:

$a=1, b=4, and, c=(-6-y)$

to get

$x=-2\pm2\sqrt{10+y}$

Nearly, it should be

$$x=-2\pm\sqrt{10+y}$$

because when you factorize the 4 out of the surd, you need to take the root of that so what you would've had was

$$x=\frac{-4\pm\sqrt{4^2+4(6+y)}}{2}$$

$$x=\frac{-4\pm\sqrt{4(4+(6+y))}}{2}$$

$$x=\frac{-4\pm\sqrt{4}\sqrt{4+(6+y)}}{2}$$

$$x=\frac{-4\pm2\sqrt{4+(6+y)}}{2}$$

At this point, would I swap x and y to get the inverse of f(x)?

Yes but the inverse needs to be a function, and you can't possibly have that with a $\pm$ there. You need to restrict the domain of your original quadratic for there to be an inverse that's 1:1.

 

[crybllrd]

OK, thank you.
I did have the quadratic right, I just typed it wrong.
I split it up into two functions from the +/-.
When I graph them, however, it does not look like it is mirrored over y=x.

EDIT: I figured it out, I didn't cancel out the 2 in the numerator.
All is well,
thanks again everyone~