# How do you decide whether the relation x = y^2 - 2y + 1 defines a function?

Nov 27, 2015

This relation is not a function.

#### Explanation:

Your relation is not a function.

The reason for this is that you can find an $x$ where $y$ doesn't have a unique value.

Let's transform the right side a bit so that it's easier to see:

$x = {\left(y - 1\right)}^{2}$

Now, if this relationship was a function, you would need to obtain one unique ("one and only one") value for $y$ for any value of $x \in \mathbb{R}$.

However, this doesn't hold here:

E.g., for $x = 5$, both $y = 6$ and $y = - 4$ provide the equality:

$5 = {\left(6 - 1\right)}^{2}$ and $5 = {\left(- 4 - 1\right)}^{2}$ are both true.

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Another way to see this is to graph the function:

graph{x = y^2 - 2y + 1 [-5, 15, -5, 5]}

Here, you also see very clearly that:

• the function is not defined for $x < 0$,
• for $x = 0$ it has indeed one unique value for $y$,
• and for $x > 0$, each $x$ has two $y$ values.