#1

x / (x(x-1))

I need to find (if any) the holes, VA, HA, Domain and Range.

I got:

Holes @ x=0

VA @ x=1

HA @ y=0

Domain = All real #s except x=0,1 (the hole and the VA)

Range = All real #s except y=0, -1 (the HA and the "guess")

My dilemma:

I know the graph crosses the x-axis and needed to find out what the “y” coordinate is at that value (the hole) in order to correctly identify the range; i guesstimated it, which is that red -1 value. After taking a wild guess, I am assuming its -1. I got by plugging in 0.1 and -0.9 into the equation and getting the values -1.111 and -0.999. I think i need to take the derivative of something. I have taken calculus before and can't put my finger on it, but know i can do something to find that value.

Anyone else have the way to correctly identify that y-value with a full proof method, rather than just guessing? (btw, no calculators are allowed)

Thanks

I need to find (if any) the holes, VA, HA, Domain and Range.

I got:

Holes @ x=0

VA @ x=1

HA @ y=0

Domain = All real #s except x=0,1 (the hole and the VA)

Range = All real #s except y=0, -1 (the HA and the "guess")

My dilemma:

I know the graph crosses the x-axis and needed to find out what the “y” coordinate is at that value (the hole) in order to correctly identify the range; i guesstimated it, which is that red -1 value. After taking a wild guess, I am assuming its -1. I got by plugging in 0.1 and -0.9 into the equation and getting the values -1.111 and -0.999. I think i need to take the derivative of something. I have taken calculus before and can't put my finger on it, but know i can do something to find that value.

Anyone else have the way to correctly identify that y-value with a full proof method, rather than just guessing? (btw, no calculators are allowed)

Thanks

#2

x / (x(x-1)) =1/(x-1) ....draw that?

#3

-1 seems right to me.

Plug -1 in for Y in the equation.

x / (x(x-1) = -1

Now, Get the top x by itself.

-x^2+x = x

Now, solve.

-x^2 = 0

Ends up being 0 = -x.

Being as how x = 0 means y = 0, it can't also mean y = -1, therefore, that would be your hole.

Plug -1 in for Y in the equation.

x / (x(x-1) = -1

Now, Get the top x by itself.

-x^2+x = x

Now, solve.

-x^2 = 0

Ends up being 0 = -x.

Being as how x = 0 means y = 0, it can't also mean y = -1, therefore, that would be your hole.

#4

What the ****?

#5

i think it is -1 man. You were basically finding the limit of f(x) as x approches zero which is how you get where the zero would be based on the x coordinate.

#6

wow im glad im only in algebra 2 this upcoming year

haha

haha

#7

i think it is -1 man. You were basically finding the limit of f(x) as x approches zero which is how you get where the zero would be based on the x coordinate.

YEA! Thats it, f(x) as X -> is the finding the derivative of my equation.

But, how do you find the derivative of the function. I think its f(x+h)-f(x) / h, but i was pretty sure there was a easier way to do it when it was in a fraction.

I know -1 is pretty much the right answer, but i need a concrete way of getting to it.

#8

Math? During summer? Blasphemy!

#9

wtf is a hole

#10

I think i just got it, or this might have been a coincidence.

The derivative of that equation is 1/ (2x-1).

Then i just plug in my x-value that i want to find, which in this case, is 0, and tada! My answer is -1! Was right in front of me the whole time.

The derivative of that equation is 1/ (2x-1).

Then i just plug in my x-value that i want to find, which in this case, is 0, and tada! My answer is -1! Was right in front of me the whole time.

#11

yeah you're right for sure

wow i forgot about all that derivative garbage - good stuff

not

wow i forgot about all that derivative garbage - good stuff

not