CALCULUS help, pleez.

[peterhuang913]

since you cannot divide by zero, you will have an asymptote where x³ -9x = 0
x³ = 9x
x² = 9
x = SQRT(9) = ± 3

asymptotes are at positive 3 and negative 3

*edit*
realised you're looking for the y intercept:

y=(2x^3-8x)/(x^3-9x)
dy/dx = [(x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)]/(x^6-81x^2)
y intercept is when dy/dx = 0

0 = [(x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)]/(x^6-81x^2)

you can ignore the denominator, since only the top needs to equal zero

0 = (x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)
0 = (6x^5-8x^3-54x^3-72x)-(6x^5-18x^3-24x^3-72x)
0 = 10x^3-30x^2

It should be easy to do from here.

*edit*
Okay, just found out something really easy:
y = (2x³-8x)/(x³-9x)
simplifies easily to
y = (2x²-8)/(x²-9)
when x = 0
y=(0-8)/(0-9)
= 8/9

It's been a little while since I've actually done any calculus, but I can still remember it

the last part has been mentioned already

 

[Apokalipse]

the last part has been mentioned already

Ah, didn't see that.
Probably because it wasn't explained algebraically (I tend to use algebra where I can to explain something mathematically)

 

[Chankama]

Y-intercept occurs when x=0.

y intercept is when dy/dx = 0

Why'd u say this apok? I can point out many situations where the y-intercept has nothing to do with the slope - dy/dx. Think of a line for example.

To find the y-intercept just set x=0. On a two dimensional (x-y) axis, that IS what the y-intercept is.

0 = [(x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)]/(x^6-81x^2)

you can ignore the denominator, since only the top needs to equal zero

Whenever you ignore the common denominator (multiplying it out with zero), in any math problem, you should always state the values of the denominator which make it zero. If the top solves to have an x-value that is restricted b/c of the denominator, you cannot say it's a potential value. Neither can you say it's not - you just need to find another way for that case.

In anycase, for this problem, there is no y-intercept. Asymptotes are not y-intercepts. That's a very different property of the curve.

 

[kfc469]

realised you're looking for the y intercept:

y=(2x^3-8x)/(x^3-9x)
dy/dx = [(x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)]/(x^6-81x^2)
y intercept is when dy/dx = 0

0 = [(x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)]/(x^6-81x^2)

you can ignore the denominator, since only the top needs to equal zero

0 = (x^3-9x)(6x^2-8)-(2x^3-8x)(3x^2-9)
0 = (6x^5-8x^3-54x^3-72x)-(6x^5-18x^3-24x^3-72x)
0 = 10x^3-30x^2

It should be easy to do from here.

I know that this works for this particular equation, but this doesn't work for all equations does it? This just finds when the slope of the graph is 0 which means that the graph has a min or max there. Just because the graph has a min or max at a point doesn't mean that that point is a y-int.

The only reason this method works for this equation is because the y-int just so happens to be at a max in the function, correct?

 

[Chankama]

I know that this works for this particular equation, but this doesn't work for all equations does it? This just finds when the slope of the graph is 0 which means that the graph has a min or max there. Just because the graph has a min or max at a point doesn't mean that that point is a y-int.

The only reason this method works for this equation is because the y-int just so happens to be at a max in the function, correct?

Correct - about the max/min.

But, this curve doesn't have a y-intercept. Don't really know who said it was 8/9. Even it IS on the back of the book ..

These books are written by high school teachers..



EDIT:

Whoever did it just divided the top and the bottom by "x" and said it's 8/9 when you set x=0. This is incorrect. WHen you divide the top and bottom by x, you HAVE to say that x != 0 as part of the resulting equation.

x/x = 1 at everywhere other than x=0

The y-intercept is not a limit. It's simply what the value of the equation is when x=0.

 

[kfc469]

Y-intercept occurs when x=0.



Why'd u say this apok? I can point out many situations where the y-intercept has nothing to do with the slope - dy/dx. Think of a line for example.

To find the y-intercept just set x=0. On a two dimensional (x-y) axis, that IS what the y-intercept is.



Whenever you ignore the common denominator (multiplying it out with zero), in any math problem, you should always state the values of the denominator which make it zero. If the top solves to have an x-value that is restricted b/c of the denominator, you cannot say it's a potential value. Neither can you say it's not - you just need to find another way for that case.

In anycase, for this problem, there is no y-intercept. Asymptotes are not y-intercepts. That's a very different property of the curve.

yeah, we said the same thing. Whenever you do away with the denominator by multiplying it by 0, you need to be sure that you include it in the domain. So it you have 0=x^2/14x, you can simplify that to x^2, but you need to be sure that you include x cannot = 0 in the domain which means that that equation has no real solutions.

 

[Apokalipse]

y intercept is when dy/dx = 0

Why'd u say this apok?

 

[Chankama]

I saw that graph. It's wrong. All graphing programs do is sample some points and connect the dots. Obviously, they can't sample "each" point on the curve. Samples are discrete. A curve is continuous.

In anycase, questions like this shouldn't be solved using graphs. It's just a simple analytical problem....

As I said before, you can't just get rid of denominators without making a note of them.

y=x/x=1, anywhere other than x=0....

i.e. what's the value of "y" if xy = x, when x = 0?

Try graphing y=x/x in that graphing program.

Only use graphs as a helpful tool to visualize and never as a solution. This curve simply does not have a y-intercept.

 

[Apokalipse]

I saw that graph. It's wrong. All graphing programs do is sample some points and connect the dots. Obviously, they can't sample "each" point on the curve. Samples are discrete. A curve is continuous.

In anycase, questions like this shouldn't be solved using graphs. It's just a simple analytical problem....

As I said before, you can't just get rid of denominators without making a note of them.

y=x/x=1, anywhere other than x=0....

i.e. what's the value of "y" if xy = x, when x = 0?

Try graphing y=x/x in that graphing program.

Only use graphs as a helpful tool to visualize and never as a solution. This curve simply does not have a y-intercept.

This is without using the graph:

y = (2x³-8x)/(x³-9x)
simplifies easily to
y = (2x²-8)/(x²-9)
when x = 0
y=(0-8)/(0-9)
= 8/9

*edit*
just saw your other post:

Whoever did it just divided the top and the bottom by "x" and said it's 8/9 when you set x=0. This is incorrect. WHen you divide the top and bottom by x, you HAVE to say that x != 0 as part of the resulting equation.

x/x = 1 at everywhere other than x=0

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

 

[joshd]

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.