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.
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".