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

Answer 1

Nathan Axel

This relation is not a function.

Your relationship isn't a system.

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

To make the right side easier to see, let's slightly alter it:

#x = (y-1)^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 RR#.

But this isn't the case here:

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

# 5 = (6-1)^2# and #5 = (-4-1)^2# are both true.

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Graphing the function is another way to visualize this:

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

It's also quite evident here that:

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Answer 2

Julia Adams

To determine if the relation x = y^2 - 2y + 1 defines a function, we need to check if each input value of y corresponds to exactly one output value of x. We can rewrite the relation as x = (y - 1)^2. Since for each value of y, there is only one corresponding value of x, the relation defines a function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.