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AR15.COM
8/29/2012 12:24:51 PM EDT
Ok, I'm helping my cousin's niece tonight, specifically x/y intercept equations. It's been a while, but IIRC you set each one to 0 and then solve the equation

if xy=3...substitute y with 0....x*0=3...0=3  Would that mean the intercept doesn't exist, would it be undefined, or am I doing something wrong?
8/29/2012 12:26:57 PM EDT
[#1]
xy=3;  therefore y=x/3

Set x=0, get y=0
set y=0, get x=0

It intercepts both the x and y axes at 0 (i.e., this equation hits the center of the target).
8/29/2012 12:27:40 PM EDT
[#2]
Quoted:
Ok, I'm helping my cousin's niece tonight, specifically x/y intercept equations. It's been a while, but IIRC you set each one to 0 and then solve the equation

if xy=3...substitute y with 0....x*0=3...0=3  Would that mean the intercept doesn't exist, would it be undefined, or am I doing something wrong?


y = 3 / x


1 = 3
3 = 1
6 = .5

right?
8/29/2012 12:34:11 PM EDT
[#3]
Linear equation.

Ergo the intercept is at (2, 288)

8/29/2012 12:35:33 PM EDT
[#4]
Quoted:
xy=3;  therefore y=x/3

Set x=0, get y=0
set y=0, get x=0

It intercepts both the x and y axes at 0 (i.e., this equation hits the center of the target).


Check your equation. Then realize you divided by zero.
8/29/2012 12:35:41 PM EDT
[#5]
Convert it to the form of the slope equation and you can see that y-intercept would be zero:

Slope equation =>   y = mx + (y intercept)

Your equation in slope eq. form would be:   y = 3/x + 0

8/29/2012 12:40:12 PM EDT
[#6]
Quoted:
Ok, I'm helping my cousin's niece tonight, specifically x/y intercept equations. It's been a while, but IIRC you set each one to 0 and then solve the equation

if xy=3...substitute y with 0....x*0=3...0=3  Would that mean the intercept doesn't exist, would it be undefined, or am I doing something wrong?


Means you have some asymptotes.
8/29/2012 12:42:48 PM EDT
[#7]
When x=0 the answer is undefined.
8/29/2012 12:45:01 PM EDT
[#8]
Quoted:
xy=3;  therefore y=x/3

Set x=0, get y=0
set y=0, get x=0

It intercepts both the x and y axes at 0 (i.e., this equation hits the center of the target).


No.

xy=3, therefore y=3/x and x=3/y

No dividing by zero allowed.  This isn't a line - maybe a hyperbola?
8/29/2012 12:47:41 PM EDT
[#9]
Quoted:
Convert it to the form of the slope equation and you can see that y-intercept would be zero:

Slope equation =>   y = mx + (y intercept)

Your equation in slope eq. form would be:   y = 3/x + 0



What do you divide 3 by to get 0?
8/29/2012 12:49:36 PM EDT
[#10]
Quoted:
Quoted:
Convert it to the form of the slope equation and you can see that y-intercept would be zero:

Slope equation =>   y = mx + (y intercept)

Your equation in slope eq. form would be:   y = 3/x + 0



What do you divide 3 by to get 0?


infinity
8/29/2012 12:50:22 PM EDT
[#11]
Oh, math. I thought it said meth.
8/29/2012 12:52:10 PM EDT
[#12]
Quoted:
Quoted:
xy=3;  therefore y=x/3

Set x=0, get y=0
set y=0, get x=0

It intercepts both the x and y axes at 0 (i.e., this equation hits the center of the target).


No.

xy=3, therefore y=3/x and x=3/y

No dividing by zero allowed.  This isn't a line - maybe a hyperbola?





8/29/2012 12:52:41 PM EDT
[#13]
Solve for y = mx + b, multiply by 87 and finally, divide by 0.
8/29/2012 12:53:51 PM EDT
[#14]
XY=3 has asymptotes along both axis which go to infinity.  Thus, there are no intercepts on the x or y axis.
8/29/2012 12:53:52 PM EDT
[#15]
Quoted:
Quoted:
Convert it to the form of the slope equation and you can see that y-intercept would be zero:

Slope equation =>   y = mx + (y intercept)

Your equation in slope eq. form would be:   y = 3/x + 0



What do you divide 3 by to get 0?


I just said the y-intercept in the slope form eq. is valueless.  Y would go to infinity if you divide 3 by 0.
8/29/2012 12:54:16 PM EDT
[#16]
Limit as X approaches zero is infinity. Limit as Y approaches zero is infinity.

Reflects about the y = -x
8/29/2012 12:58:49 PM EDT
[#17]
Quoted:
When x=0 the answer is undefined.


I'm leaning toward this, but just to make it interesting lets add some exponents


x^2 + y^2 = 1....i'm thinking (-1, 0) (1,0) and (0,-1) (0, 1)
8/29/2012 1:08:21 PM EDT
[#18]



Quoted:



Quoted:

When x=0 the answer is undefined.




I'm leaning toward this, but just to make it interesting lets add some exponents





x^2 + y^2 = 1....i'm thinking (-1, 0) (1,0) and (0,-1) (0, 1)
Yup, that's the unit circle.





 
8/29/2012 1:16:23 PM EDT
[#19]
Quoted:
Linear equation.

Ergo the intercept is at (2, 288)



This gets my vote.



8/29/2012 1:20:06 PM EDT
[#20]
Quoted:

Quoted:
Quoted:
When x=0 the answer is undefined.


I'm leaning toward this, but just to make it interesting lets add some exponents


x^2 + y^2 = 1....i'm thinking (-1, 0) (1,0) and (0,-1) (0, 1)
Yup, that's the unit circle.

 


Indeed it is.
8/29/2012 1:39:46 PM EDT
[#21]


8/29/2012 2:07:09 PM EDT
[#22]
Quoted:
Quoted:
Convert it to the form of the slope equation and you can see that y-intercept would be zero:

Slope equation =>   y = mx + (y intercept)

Your equation in slope eq. form would be:   y = 3/x + 0



What do you divide 3 by to get 0?


Chuck Norris can do that.
8/29/2012 2:20:14 PM EDT
[#23]
Quoted:
XY=3 has asymptotes along both axis which go to infinity.  Thus, there are no intercepts on the x or y axis.


This is the correct answer.