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Determine whether the equation $y^2=x+6$ defines $y$ as a function of $x$. Find its domain and range.

What I have tried: $$y^2=(\sqrt{x+6})^2\Longrightarrow y=\pm\sqrt{x+6}$$

Therefore, the above equation does not represent a function, because to be a function, for each input there must be a unique output. But here there is no unique output for any $x>-6$.

But for domain and range, can we say that domain is $[-6,\infty )$ and range is $\mathbb{R}$? I ask because I know we can plot it in Cartesian coordinates.

Thank you.

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  • $\begingroup$ Yes it is not a function. So we can not find its,domain and range. But my question is,when we plot in coordinate axis we get real values,of $x$ and,$y$.whats wrong with it $\endgroup$ Commented Apr 21, 2020 at 20:37
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    $\begingroup$ In this case, DO NOT call it "a function", but DO call it "$y^2=x+6$ correspondence". In $x,y$-correspondences, you can find domains and ranges as you did. You were right. $\endgroup$ Commented Apr 21, 2020 at 20:49

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Your explanation is correct, but try to be more rigorous in your answers. For example, for the domain: $$y^2 >0 \implies x > -6 \Leftrightarrow x \in \left[-6,\infty\right).$$

Another way to see that $y$ does not define a function, is by indentifying the equation of parabola.

More specifically, The pencil of conic sections with the $x$ axis as axis of symmetry, one vertex at the origin $(0, 0)$ and the same semi-latus rectum $p$ can be represented by the equation:

$$y^2 = 2px + (e^2-1)x^2, \quad e \geq 0.$$

By your expression, the eccentricity is $e=1$ and $p=1/2$. Then, you can just shift this parabola by $+6$ and it maintains the same properties (meaning that $x$ will be the axis of symmetry).

Being able to plot something in cartesian coordinates doesn't mean that it represent a function, as you saw yourself. The definition of a function is strict and that answers your comment question "what's wrong with it".

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  • $\begingroup$ So what would be the final answer for domain and range of this question. It does not exists because it does not represent function $\endgroup$ Commented Apr 21, 2020 at 20:41
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    $\begingroup$ Every expression has a domain in which it can be defined. Also every expression has a range. These does not fade away as terms just because you do not have a function. For example, your given expression is defined only when $x > -6$. This is a domain. Now, you may not call this a function domain of course. Same goes for $y$ and the range of values it can attain. $\endgroup$ Commented Apr 21, 2020 at 20:44

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