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Mathematics 16 Online
OpenStudy (anonymous):

what is the inverse of y=2x^2 +8x+3 ?? help me please!

OpenStudy (agent0smith):

It won't have a real inverse, since an x^2 function is not invertible as it's not one-to-one.

OpenStudy (agent0smith):

You can invert it by completing the square (this is also how the quadratic formula is derived)

OpenStudy (anonymous):

i am confused can you give me an example?

OpenStudy (agent0smith):

But it's still not a technical inverse, since parabolas are not one-to-one functions

OpenStudy (anonymous):

um.. what if the function is restricted?

OpenStudy (anonymous):

to one to one function

OpenStudy (agent0smith):

Yes, you can restrict it to one-to-one. To complete the square: \[y=2x^2 +8x+3 \] \[y=2(x^2 +4x)+3\] subtract 8 in this next step, since we added 8 from the parentheses: \[y=2(x +2)^2+3-8\] \[y=2(x +2)^2-5\]

OpenStudy (agent0smith):

Now try rearranging that to solve for x.

OpenStudy (anonymous):

I see what is happening now! thank you :)

OpenStudy (agent0smith):

Good :) \[y=2(x +2)^2-5 \] just rearrange it until you have x = ....\[y+5=2(x +2)^2\]

OpenStudy (anonymous):

yeh!! it is clearer thank you so much for your help :)

OpenStudy (agent0smith):

And then swap the y with x and x with y.

OpenStudy (agent0smith):

@rizwan_uet wolframalphas inverse appears to be wrong... looks like it should be \[-2 ± \sqrt \frac{ x+5 }{ 2 }\]

OpenStudy (agent0smith):

Their graph is correct, but the equation wolfram has is not.

OpenStudy (anonymous):

@agent0smith wow

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