[WorldIndustries]

Studying for UP Calculus, and I'm getting confused by some stuff...

Okay, so this specific type of funtion follows this pattern (example):

(2x³ - 8x) / (x³ - 9x)

I know how to find the horizontal and vertical asymptotes, the point of discrepancy, the x-intercepts... But where the **** is the y-intercept? I've got 8/9 written as the correct answer, but I don't know where to find it... I know the x terms in the denominator/numerator would give me 8/9, but that can't be right...

Anyone?


PS: If you want to SEE the graph, go to <a href=http://www.walterzorn.com/grapher/grapher_e.htm>this</a> page, and paste "(2x^3-8x)/(x^3-9x)" into the function box...

[Vybuni]

That graph has no y-intercept. To find the y-intercept, plug in 0 for all the x terms. In that equation, you get 0/0, which is undefined, giving you a hole at x=0 (the y-intercept).

The limit as x approaches zero, however, is 8/9. To find that, factor out and cancel an x from the top and bottom and plug in 0 for x, which gives you 8/9.

[Mr. Switch]

I remember doing those in algebra trig back in the day..






BAD MEMORIES.

[kfc469]

Do you have a graphing calculator? If not, I would strongly suggest getting one. I have a TI-89 and it does everything for me. It takes derivatives, finds limits, solves equations with an unlimited number of variables, etc.

[WorldIndustries]

Yes, I have a graphing calculator... I know what the answer is... But the thing is, I needed to know how to get that answer on paper. You kinda need to do that on the test, to show you know how to get the answer.

Eleventeen year olds would be able to do this stuff if you just had to plug it into the calculator...

[peterhuang913]

one way is to use a graphing calculator and find its relative maximum

[kfc469]

Originally Posted by WorldIndustries

Eleventeen year olds would be able to do this stuff if you just had to plug it into the calculator...

Sorry, I've never heard of one of those.

Originally Posted by peterhuang913

one way is to use a graphing calculator and find its relative maximum

First, you would have to find the rel min/max of the derivative, not the original function and that would give you the x intercepts. Second, he wants the y ints.

[peterhuang913]

Originally Posted by kfc469

Sorry, I've never heard of one of those.


First, you would have to find the rel min/max of the derivative, not the original function and that would give you the x intercepts. Second, he wants the y ints.

did you check the graph, the relative max is the bump in the center

Originally Posted by Vybuni

The limit as x approaches zero, however, is 8/9. To find that, factor out and cancel an x from the top and bottom and plug in 0 for x, which gives you 8/9.

now this is a good answer. not limits and derivatives

Originally Posted by kfc469

Sorry, I've never heard of one of those.

Eleventeen - Wikipedia, the free encyclopedia i think its a band

[kfc469]

Originally Posted by peterhuang913

did you check the graph, the relative max is the bump in the center

I know what a rel max is. Thats not what he was looking for. He was looking for the y-int's.

EDIT: I think if figured out what you were trying to say. For this particular graph, it does appear that the rel max and the y int are the same, but this is not the case from all functions.

[Apokalipse]

Originally Posted by WorldIndustries

Studying for UP Calculus, and I'm getting confused by some stuff...

Okay, so this specific type of funtion follows this pattern (example):

(2x³ - 8x) / (x³ - 9x)

I know how to find the horizontal and vertical asymptotes, the point of discrepancy, the x-intercepts... But where the **** is the y-intercept? I've got 8/9 written as the correct answer, but I don't know where to find it... I know the x terms in the denominator/numerator would give me 8/9, but that can't be right...

Anyone?


PS: If you want to SEE the graph, go to this page, and paste "(2x^3-8x)/(x^3-9x)" into the function box...

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