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VCE Specialist 3/4 Question Thread!

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^^^111^^^:

--- Quote from: mabajas76 on December 29, 2021, 01:43:04 pm ---The math symbols wern't working for me so i will refer to the 1/2x^2-4 as f(x).
So f(x)=1/0, that is how you get the asymtopes right? And then for the reciporcal f(x)^-1, the vertical asymtopes become the x intercepts as you know. Now, when you take the reciporcal of one side, you do the same to the other. so F(x)^-1=0/1=0. So there would not be holes because you can't find a value for which f(x)^-1=1/0, because the reciporcal f(x) has a value = to 1/0 and the new function is a reciporcal of this.

Hope that helped haha, sorry if it didn't.

--- End quote ---

Ohh ok,

so just to confirm I'm understanding this, since we are taking the reciprocal, it would be to both sides, so that means it's 1/y = 1/f(x) or y = f(x). So there won't be holes, as this would be 0/1 (0 is numerator and 1 is denominator), instead of 1/ (1/0) (for example).

Thank you!

mabajas76:
Well when you have f(x), that isn't an asymtope right? It is only when x= + or - sqrt of 2 that it becomes an asymptope, because + or - sqrt of 2 is equal to 1/0, so with 1/f(x), the y value for any x will become 1/y. So thats why at sqrt 2 it goes from 1/0(undefined) to 0/1.

Now what ur saying is if 1/(1/0) has an asymtope, I got what ur saying because its like saying 1/undefined which doesn't make sense. or 1/f(x), but f(x) has values which are undefined. Well the thing is undefined can be exprezsed as 1/0, that is its definition in numbers, a division by 0. And 1/(1/0) will actually give it a value. So no there aren't any values that 1/f(x) would still have asymtopes because even though the f(x) can still give a 1/0, it isn't ready to be evaluatated because its divided by 1.

galaxysauce:
Hey everyone!

Can someone please help me with this proof question? Thank you!  :)

fun_jirachi:
Recall that \[\cos \theta = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}||||\vec{v}||}.\]

Choose some arbitrary vector \(\vec{v} = (x_1, y_1, z_1)\). Then, we can say that it makes an angle of \(\alpha\) with the unit vector \((1, 0, 0)\).

Then,
\[\cos \alpha = \frac{x_1}{\sqrt{x_1^2+y_1^2+z_1^2}} \implies \cos^2 \alpha = \frac{x_1^2}{x_1^2 + y_1^2+z_1^2}.\]

The rest of the proof follows by doing the same thing on the other coordinate axes then summing the results up.

galaxysauce:
Hi.. could someone please help me with this question? It's a pretty simple one, but I'm stuck on drawing the "other" triangle, and use it to find the other "BC".

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