Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle

Question

So why is it a function, even though for example $x = 8$; you'll have $y = +2$ and $y = -2$. It'll fail the vertical line test. But every textbook considers it as a function. Did I misunderstand something?

Edit: Wait how come $y = \sqrt{4 - x^2}$ is a function too when it can be transformed into $$ y = \sqrt{4-x^2} $$ $$ y^2 = 4- x^2$$ $$ x^2 + y^2 = 4 $$ $$ \frac{x^2}{4} + \frac{y^2}{4} = 1$$

which is an equation of a circle and not a function

Answer 1

Consider these two questions:

1. Solve for $y$ in the equation: $y^2 = 4$.
2. Evaluate $\sqrt{4}$.

These questions are related, but they are not the same.

For the first question, there are two answers: $\pm 2$, since both of these numbers squared will give you $4$.

The second asks: "what nonnegative number, when squared, gives you $4$?" The answer to that is $2$ (and not $-2$). In general for a nonnegative real number $x$, its square root, $\sqrt{x}$ is defined to be the nonnegative solution $y$ to $y^2 = x$.

Thus, $\sqrt{8-4} \neq -2$, so the graph $y=\sqrt{x-4}$ contains the point $(x,y) = (8,2)$ but not $(x, y) = (8, -2)$. In general there is at most one answer to the question "evaluate $\sqrt{x-4}$," so the graph does indeed pass the vertical line test. (You can see a sketch of the graph on Wolfram Alpha.)

(Things get more interesting when we try to define a square root of a complex number!)

Edit (in response to follow up question about $y = \sqrt{4-x^2}$): Consider the graphs of $y^2 = x$ and $y = \sqrt{x}$. Based on the discussion above, they are not the same graph. The graph $y^2 = x$ "transforms" into $y = \pm \sqrt{x}$, which is not a function.

Answer 2

It is a function because that square root sign in your expression denotes the principal square root. The principal square root of a positive number is defined as the positive square root of that number i.e. we discard the negative square root.

If that square root symbol instead denoted both the negative and positive square roots, then yes, it wouldn't be a function but a relation.

Answer 3

For the followup question:

$y = \sqrt{x}$ and $y^2 = x$ are not equivalent statements.

If $y = \sqrt{x}$ then $y^2 = x$ but not the other way around. Hence your equations $$y = \sqrt{4-x^2}$$ and $$x^2 + y^2 = 4$$ may very well have different sets of solutions, and they do. For the first of these equations, $y$ is automatically positive (because of the other discussions about the square root). The solutions of $y = \sqrt{4-x^2}$ form a semi-circle, the part of the circle $x^2 + y^2 = 4$ which correspond to points where $y \ge 0$.

Answer 4

To add some additional detail to @mrf's response, I think it is important to understand an equivalent way of interpreting the phrase:

If $y=\sqrt{4-x^2}$, then $y^2=4-x^2$ $ \quad(\dagger_1)$.

Consider the two different sets of ordered pairs $R_1$ and $R_2$ defined as:

$R_1 = \left\{ \langle x,y \rangle \in \mathbb R \times \mathbb R: y=\sqrt{4-x^2} \right\}$

$R_2 = \left\{ \langle x,y \rangle \in \mathbb R \times \mathbb R: y^2=4-x^2 \right \}$

Another way of interpreting $(\dagger_1)$ is:

If $\langle x,y \rangle \in R_1$, then $\langle x,y \rangle \in R_2 \quad (\dagger_2)$

We can see that all of the ordered pairs in $R_2$, when plotted on a cartesian coordinate system, form a circle; $R_1$ is the upper half of this circle. This is a visual way of saying that $R_1 \subset R_2$, which is why $(\dagger_2)$ is valid.

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Comparatively, consider the statement:

If $y^2=4-x^2$, then $y = \sqrt{4-x^2} \quad (\dagger_3)$

To understand why this statement is false, note that $(\dagger_3)$ amounts to saying that:

If $\langle x,y \rangle \in R_2$, then $\langle x,y \rangle \in R_1$.

Look at the graph of $R_2$. Choose an ordered pair that is on the lower half of the circle. Clearly, this ordered pair is not on the upper half of the circle...i.e. this ordered pair cannot be in $R_1$.

Hence, $(\dagger_3)$ is false.