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Find the equation of the inverse of y = (x + 2)^2 - 4

Edit: Simply switching around the x and y doesn't work because then you are stuck with solving for y

x = y^2 + 4y

There should be a simpler method

Edit: The domain is restricted to x is greater than or equal to -2

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  • $\begingroup$ This function is not injective. What do you mean by "inverse"? $\endgroup$ – ajotatxe Dec 26 '16 at 23:16
  • $\begingroup$ Sorry, i forgot to mention the restriction $\endgroup$ – TripleA Dec 26 '16 at 23:17
  • $\begingroup$ Hint: $x+2=\pm \sqrt{y+4}$ but since $x+2 \ge 0 \,\cdots$ $\endgroup$ – dxiv Dec 26 '16 at 23:20
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Strictly speaking, this function does not have an inverse, since it's not injective (i.e. it's not "one to one", which is to say that two domain elements can map to the same element in the range). However, if we restrict the domain to $x\geq -2$, we can invert the function. To do this, just swap $x$ and $y$ and solve for $y$: $$0 = y^2+4y-x$$ using the quadratic formula, $$y = \frac{-4 +\sqrt{16+4x}}{2} = -2+\sqrt{4+x}$$

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