Question

4x=y^2 determine if the equation defines y as a function of x

Answer 1

i have no clue bout this

Answer 2

Suppose \(x=1\). Then you have
\[4=y^2\]
There are two possible solutions here: \(y=2\) and \(y=-2\).

Since there are two values of \(y\) for one value of \(x\), \(y\) is not a function of \(x\).

Answer 3

y did we suppose x is one?

Answer 4

can we suppose x is any number but keep it as 1 to keep it easy?

Answer 5

You could really let \(x\) be anything, within reason.

In general, \(x\) can't be negative because \(y^2>0\) for any \(y\).

So you're left with \(x\ge0\).

If you let \(x=0\), then you have \(0=y^2~\Rightarrow~y=0\) (one \(y\) for one \(x\)).

But if you let \(x>0\), then you have \(x=y^2~\Rightarrow~y=\pm\sqrt x\) (one \(y\) for more than one \(x\)).

And yes, I chose 1 for convenience.

Answer 6

thanx buddy

Answer 7

You're welcome!

Answer 8

see ya later!!!

Answer 9

Bye