I don't know what to do with this problem. It's 3x-y^2=16 It wants to know if it's a function. I know what a function is, and how to figure out if an equation is a function, and why or why it may not be, but this one stumps me.
of course it is not a function it is an equation
-y^2=16-3x y^2=3x-16 y=sqrt(3x-16)
the real question is, if you solve for \(y\) what do you get lets do it carefully
this is a function @satellite73
ur no
oh yeah
\[ 3x-y^2=16 \]is an equation, it is not a function the person who wrote this question is confused, but what should really be asked is can \(y\) be function of \(x\) and the answer is NO
\[3x-y^2=16\\ 3x-16=y^2\\ \pm\sqrt{3x-16}=y\] and the \(\pm\) tells you x is not a function of y
or rather y is not a function of x
Oh wait, I'm sorry, I forgot the 4. The equation is actually 3x-4y^2=16. @satellite73 Equations are functions. You plug in x values to get y values. An equation is a function if only one y value comes out per x value.
right the 4 makes no difference the answer is still NO because of the square on the y
The 4 makes a difference in the equation, though. So, what does the equation turn out to be when solving for y?
ahhh thus begins the confusion between equations and functions functions CAN be given by equations, but they do not have to be and all equations are not functions
\[ 3x-4y^2=16. \\ 3x-16=4y^2\\ \frac{3x-16}{4}=y^2\\ \pm\frac{\sqrt{3x-16}}{2}=y\]
I know that, I didn't say all equations were, I just said that equations can be functions, too.
in any case it is the \(\pm\) that throws it off be careful when solving with a square common mistake would be to write \[y=\frac{\sqrt{3x-16}}{2}\] and forget about the \(\pm\) part
another way to see it is not a function is to know that this is a parabola that opens to the right, so does not pass the vertical line test http://www.wolframalpha.com/input/?i=3x-y^2%3D16+
So, the question still remains, is this a function. Does every x input have only one y output?
and the answer is NO we can do it by example if you like
can we find two different y values that come from the same x value in other words, and the answer is yes we can pick an x and we can find the two y values
I graphed it on my graphing calculator, and got half a sideways parabola, meaning it would pass the vertical line test.
did you get something like this ?|dw:1418787130484:dw|
No, like I just said, I got half a sideways parabola.
that is because your graphing calculator only graphs functions, so it probably graphed the top half only
like i said, the square on the y is what makes it not a function just like \(x=y^2\) is not a function since if \(x=25\) \(y=5\) or \(y=-5\) both work
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