bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
tkhunny said:That is not difficult to typoe in.
f(x) = (x+3)/(x-5)
Okay, what is it that you want to know. There are two asymptotes, one horizontal and one veritcal.
What is your plan for identifying them?
You can also learn just a little LaTeX
\(\displaystyle f(x) = \frac{x+3}{x-5}\)
tkhunny said:bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
Not tricky at all. Factor and read. You shoudl recognize a "Difference of Squares".
Okay, quit trying to get me to do all your homework. You show me the next one.
tkhunny said:It's easy to find. Here's one place. http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/
You'll have a spot of trouble completing the square on that cubic expression.
It has one Positive Real zero. Can you find it?
The numerator has three Real zeros. Can you find those?
Really, you must find only one. After taht, you can reduce to a quadratic and those are easy, right?
Horizontal Asymptote is an eyeball problem. QUICK - Write it down!
bcddd214 said:tkhunny said:bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
Not tricky at all. Factor and read. You shoudl recognize a "Difference of Squares".
Okay, quit trying to get me to do all your homework. You show me the next one.
correct?
f(1)=(3x-1)/?(x^2-1)=(3(1)-1)/?(1^2-1)=2/0 is a VA
f(-1)=(3x-1)/?(x^2-1)=(3(-1)-1)/?(?-1?^2-1)=-4/0 is a VA
tkhunny said:It's easy to find. Here's one place. http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/
You'll have a spot of trouble completing the square on that cubic expression.
It has one Positive Real zero. Can you find it?
The numerator has three Real zeros. Can you find those?
Really, you must find only one. After taht, you can reduce to a quadratic and those are easy, right?
Horizontal Asymptote is an eyeball problem. QUICK - Write it down!
bcddd214 said:What I got
f(5)=(5+3)/(5-5)=8/0 is a VA
bcddd214 said:[correct?
f(1)=(3x-1)/?(x^2-1)=(3(1)-1)/?(1^2-1)=2/0 is a VA
f(-1)=(3x-1)/?(x^2-1)=(3(-1)-1)/?(?-1?^2-1)=-4/0 is a VA
What do you mean, "what do you mean?" It's pretty clear, isn't it? Youshoudl be able to spot it and write it's equatino - instantly!bcddd214 said:What do you mean "Horizontal Asymptote is an eyeball problem. QUICK - Write it down!"?
The ONLY way for me to know that is true is by you showing your work. Without much effort, I must assume you are nowhere near the prerequisite coursework,bcddd214 said:Not trying to get you to do my homework, leading step and a check of the final will be just fine.
tkhunny said:bcddd214 said:What I got
f(5)=(5+3)/(5-5)=8/0 is a VA
That makes no sense.
1) Never EVER put inappropriate values in a funcction. x = 5 is not in the Domain. You can't do that.
2) x = 5 is the vertical asmptote.