so im taking calculus AB and i got a takehome quiz, here are the problems im having trouble with and my (questionable) answers

please help, and thank you in advance


1 find vertical asymptotes of f(t) = t / (4t^2 - 2t -12)

i factored the denominator into 2(t - 2)(2t + 3)
normally would then say there asymptotes at x = 2 and x = -3/2

does that 2 on the outside effect my end result at all?


2 more to come

Your right, those are the asymptotes. Here is why.

the function is a fraction. fractions can not have 0 for the denominator. So the places where there will be no value for f{t} (cuz it would be undefined because of division by zero) is:

4t^2 - 2t - 12 = 0

2(t-2)(2t+3)=0
(divide both sides by 2)
(t-2)(2t+3)=0
you know what to do from here.

But that is why that "2" didn't matter cuz you can get rid of it by dividing it out (0/2 = 0).

Try plugging those numbers back into original equation to see if the 2 mattered. It won't increase the number of zeroes, only could possibly scale. In this case I think you are fine

ty ty

ok 2

determine the limit of x as it approaches 2 (see if it is positive or negative infinity)

function is 3 / (x-2)^2

by putting it into my calculator i see that it approaches positive infinity, but my teacher expects us to be able to solve everything without one. i dont see how to do this any other way (possibly plugging in values close to 2? such as 1.999 and 2.001?)

3

find limit as x approaches infinity of (6x^2-3x+1) / (-2x^2+4x-7)
im clueless for these ones..

also limit at infin of (6 - 2/x^2)

This post was edited by known954 on Sep 17 2013 04:55pm

2
that is exactly what you need to do

This post was edited by Azrad on Sep 17 2013 04:56pm

first thing, try to factor it and cancel some stuff, but that didn't work for me, so moving right along..

when x goes to infinity the numerator will be totally dominated by 6x^2 and not the other terms so lets discard them, the same for the denominator (-2x^2 will dominate)
sooo..

6x^2/-2x^2 = -3, there is your limit




That one is a little hard to read in the format you put it... if it is:
6 - {2/(x^2)}
the limit as x approaches infinity of 2/(x^2) = 0, leaving us with
6-0=6 and that is your limit

This post was edited by Azrad on Sep 17 2013 05:04pm

thank you. first one makes sense now

second one im still a bit confused

it is written exactly on my quiz as lim x-> infinity ( 6 - 2/x^2 ) *the parenthesis were there i didnt add them.

i understand that the limit of a constant is that constant

however why is 2/x^2 = 0? is a constant divided by infinity always zero, or ?

actually no need to use your calculator at all since (x-2)^2 will always be positive or zero (hint: it's a square)



yes


This post was edited by brmv on Sep 17 2013 05:55pm

the 3 above it doesnt affect that?