anonymous
  • anonymous
Looking at the rational function f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity Bella says that the function is defined at x = –1, x = 2, and x = 4. Edward says that the function is undefined at those x values. Who is correct? Justify your reasoning.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
writing equations out in English makes it very hard to read.
phi
  • phi
can you use the equation editor?
anonymous
  • anonymous
how do I do that?

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phi
  • phi
click on the equation button in the lower left of the input area
anonymous
  • anonymous
\[f(x)=(x-1)(x+2)(x+4)/ (x+1)(x-2)(x-4)\]
anonymous
  • anonymous
I couldn't find the "divide" button hahah
phi
  • phi
the fraction button is there, last row, on the right
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phi
  • phi
anyway, you hopefully know we do not allow "divide by 0" It is not done. in your problem, if you make x certain values, the "bottom" will have a zero and zero times anything is zero and you will have stuff/0 and we don't allow that.
anonymous
  • anonymous
@peachpi
anonymous
  • anonymous
Who would be right??
phi
  • phi
Do you understand that if you have (for example) (x+1) in the bottom what number (for x) makes x+1 be 0 ?
anonymous
  • anonymous
-1
phi
  • phi
and if x=-1 we will get in the bottom 0 * (-1-2)*(-1-4) or 0 * -3 * -5 notice the -3 and -5 don't matter, because when we multiply it out we get 0 *0 in the bottom* is not allowed
phi
  • phi
to prevent the divide by 0, we do not allow x=-1 then what is the answer for that expression when x=-1? answer: we don't have an answer. It is undefined. notice there are two other values that make the bottom 0 we don't allow those either. can you take a guess as to who is right for this question?
anonymous
  • anonymous
edward!
anonymous
  • anonymous
Yes?
phi
  • phi
yes.
anonymous
  • anonymous
Thanks<3

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