Help with Algebra Homework: Range of y = (x^2 - 4)/(x^2 - x - 12)

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Homework Help Overview

The discussion revolves around finding the range of the function y = (x^2 - 4)/(x^2 - x - 12). Participants are exploring concepts related to algebra and calculus, particularly focusing on the behavior of rational functions and their asymptotes.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss factoring the numerator and denominator to analyze the function's behavior. There are attempts to identify horizontal asymptotes and critical points, along with questions about how to determine the range without a calculator.

Discussion Status

The discussion includes various attempts to clarify the problem, with some participants offering hints and references to external resources. There is an ongoing exploration of critical points and their implications for the function's range, but no consensus has been reached on the final outcome.

Contextual Notes

Some participants express frustration over the complexity of the problem and the requirement to show all work without using a calculator. There are mentions of specific values and intervals, but the exact range remains unclear.

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Homework Statement



I'm stuck on this problem please help me

find the range of this equation

y = (x^2 - 4)/(x^2 - x - 12)

I know that it is [1/3,-infintiy]U[?,infinity]
yes I'm having trouble with that ?
At first I was like ok there should be a horizontal asymptote at y=1 there is in calculus deffinition the image as x approaches infinity from both sides right? or is just infinity... but you it appraoches one from both sides but it corsses one at x=-8 check the graph so how do I find how low it goes at this section also I can't use a calculator well I can but am suppose to show all work without using one i.e. have one to use not suppose to use it thanks

Homework Equations





The Attempt at a Solution

 
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Y = (x²-4)/(x²-x-12)

If you factor out the top and bottom of the equation you end up with

(x-2)(x+2) //Use the "difference of squares formula"
----------
(x+3)(x-4)

correct?

With no further simplifying to do, the only course of action you can really take is to figure out what Y cannot be right? What would the denominator have to equal, for the equation to have no solution?
 
-3 and 4... but that wasn't my question I was asking for the range... please help me with that
 
well I read them still can't solve the problem :(
 
apparently I do this I don't understand what this guy is saying... or were he got .337 I got 1/3...

(-∞ , 0.337] U [ 0.969,∞)...look at the critical values and evaluate the function as well as doing the VA consideration ...and you certainly will need a numerical calculator to find the critical values and function values
 
can someone please show me the calc to do this
 
QuarkCharmer gave you a hint. The numerator factors into (x+3)(x-4). So your function goes to infinity at x=(-3) and x=4. Use calculus to figure out what happens in the intervals (-infinity,-3), (-3,4) and (4,infinity) and put them together to get the total range.
 
Can you please help me with this part... I have to use the chain rule right and the product rule to find the critical values correct then what do I do?
 
  • #10
I think you just need the quotient rule to differentiate (x²-4)/(x²-x-12). Simplify it, set it equal to zero and solve for the critical points.
 
  • #11
what do I do after finding the critical points? Sorry i forgot how to do this stuff I should be able to find them eaisly but I don't remeber what to do after I find them...
 
  • #12
It is kind of disappointing you don't know. The critical points are candidates for being maxes or mins on each interval. What you are basically doing is sketching a graph of y. Do that. Find the value at the critical points, find the limiting behavior at each side of x=-3 and x=4, and put that together with what you already know, that y=1 is a horizontal asymptote. Now draw a nice sketch of the function. Deduce the range from that.
 
  • #13
Yeah, super disappointing.

After finding critical points, you could use a sign chart to determine which are maxima and which are minima
 
  • #14
ok well I posted this in the calc section sorry I posted it here didn't realize it would invovle calculus but you I found the cirtical values still couldn't solve the problem though even with a sign chart... please help me
 
  • #15
If you know the critical points you can sketch the function. Find out how it behaves around -3 and 4. Where is it positive and where is it negative?

ehild
 
  • #16
I found them it didn't do me anything check other topic
 

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