CALCULUS help, pleez.

[Chankama]

This is without using the graph:

*edit*
just saw your other post:
So the graph should have a hollow circle at y=8/9, where x=0, since it can't actually get to that specific point

Exactly .. That curve does not have a y-intercept.

That is what we are being taught in AS Further Maths - divide everything by the highest power of x you can, then set x to 0.

Don't really know what AS Further Maths is. But ask you teacher to solve this equation for y when x=0:

xy = x^2

Is y=0?... or maybe it's -5, or maybe 10000??

That's what bothers me about high school teachers. They don't think about the questions they are doing. It's all a procedure for them..

 

[veg1992]

all i no about calulus is that u can take the first derivative of the SA of a Sphere and turn it into the Volume of the Sphere

i won't be taking ap calc anyways it looks like i'll be taking Honors Calc and Trigonometry no!!!! stupid ap geometry, y did u give me a 71!!!!!!!!!! ah!!!

 

[Apokalipse]

Exactly .. That curve does not have a y-intercept.



Don't really know what AS Further Maths is. But ask you teacher to solve this equation for y when x=0:

xy = x^2

Is y=0?... or maybe it's -5, or maybe 10000??

How about i?
Even an imaginary number works in that equation.

 

[joshd]

Don't really know what AS Further Maths is. But ask you teacher to solve this equation for y when x=0:

xy = x^2

Is y=0?... or maybe it's -5, or maybe 10000??

That's what bothers me about high school teachers. They don't think about the questions they are doing. It's all a procedure for them..

You take A levels in the UK age 17-18. AS is the first year, and A2 is the second.

I'm not sure, I think we are supposed to check for stuff like dividing by 0 before you divide everything by x... Can't quite remember. I didn't write it down in my notes, as I seem to remember thinking "this is a bad way, I will just use my way"

 

[kfc469]

Apok, your method doesn't work for all equations. Take y=-(x+2)^2
If you take the derivative, you get: dy/dx=-2(x+2)
Then set it equal to zero and you get:
0=-2(x+2)
0=x+2
x=-2 which is not the y-int of this equation.

In the original problem of this thread, it is just a coincidence that the "apparent" y-int happened to be at the max on the graph. This is what I think that Chankama has been trying to say.

Note that I know that there is no y-int for that equation, I am just using it as an example.

 

[Apokalipse]

Apok, your method doesn't work for all equations. Take y=-(x+2)^2
If you take the derivative, you get: dy/dx=-2(x+2)
Then set it equal to zero and you get:
0=-2(x+2)
0=x+2
x=-2 which is not the y-int of this equation.

See my other post. I looked at the graph, and it appeared as though it has a zero gradient at the y intercept. That's why I used that method for this particular equation.
though later as Chankama pointed out, that point is actually nonexistent, since you cannot divide by zero.

If you look at the graph for y=-(x+2)^2, the gradient is not zero at the y intercept, so I would not use that method for this equation.