Does anyone wanna tell me why tf ln(x-1)/(x^2-4) is nothing like ln(3x)/(x^2-4), i just went through all of my professors steps and it was nothing like what I thought it should have been. (I attached his steps)
#Math 1 question pissed me off
jaunty surge
heavy tulipBOT
sudden tundra
Does anyone wanna tell me why tf ln(x-1)/(x^2-4) is nothing like ln(3x)/(x^2-4),...
what do you mean - they are 2 completely different functions
of course they're not the same
jaunty surge
Yes but I just got confused understanding the behavior of them from the face value, the fact that one has two 2 vertical asymptotes and another only one (where it is so basic, it just has a hole at 2)
I got confused because I expected there to be a horizontal asymptote at 2
sudden tundra
what about the functions x/(x^2-4) and (x-2)/(x^2-4)
do you understand why the first one has 2 vertical asymptotes and the second one only has one
jaunty surge
It can simplift cant it?
jaunty surge
do you understand why the first one has 2 vertical asymptotes and the second one...
No
sudden tundra
because when x=2 and x=-2 on the first one you get a nonzero number divided by 0
which always goes to infinity
but on the second one when x=2 you get 0/0 which doesn't necesarily approach infinity
0/0 can approach anything because really, anything times 0 is 0
same reason with ln(x-1)(x^2-4) when x=0 you get 0/0
because ln(2-1) = ln1 = 0
jaunty surge
Should i take that as a rule for future reference? (Value / zero = asymptote - zero / zero = either asymptote or hole??)
sudden tundra
in fact you can find that ln(x-1)/(x^2-4) approaches 1/4 as x goes to 2
jaunty surge
in fact you can find that ln(x-1)/(x^2-4) approaches 1/4 as x goes to 2
Using lhopitals rule?
sudden tundra
Should i take that as a rule for future reference? (Value / zero = asymptote - z...
you mean nonzero value for the first?
yes
sudden tundra
Using lhopitals rule?
yes
jaunty surge
you mean nonzero value for the first?
Ok great
I don't know what type of equations he will spring up on me in the test
sudden tundra
just know that if you get 0/0 that doesn't tell you anything about how the curve will look
until you do more math
jaunty surge
What should I do at that point?
sudden tundra
use l'hopital's rule
to find where the hole should be
sometimes it will still be an asymptote
if it's a rational function just cancel out all the common factors until there's no more 0/0
you will either be left with just 0 on the top (a hole with y coordinate 0), 0 on the bottom (vertical asymtote), or no zero on the top or bottom (a hole with nonzero y coordinate)
and you should probably mark the hole with an open circle. In reality, the hole is infinitely small, as there is only one value out of infinitely many that it is undefined
but i asume whoever your professor is wants to see that you know that there is a hole there
jaunty surge
Well this was a practice question of my own but I dont know he might
Thank you btw
It's just a confusing process getting to the shape of the graph of ln(x-1)/(x^2-4)
sudden tundra
just mark down all the asymptotes and holes
then start plugging in x values to get some more points
theres not really any easier way
jaunty surge
For that equation I knew knew that there was an asymptote at 1 because that was the domain restriction, as well as knowing that the domain did not include the value 2 towards infinity
Not knowing it is a hole, would I just solve for the limit at 2?
To determine if it was an asymptote or hole