Im confused can someone help me?

- anonymous

Im confused can someone help me?

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- schrodinger

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- anonymous

@Michele_Laino i dont understand how to solve these?

##### 1 Attachment

- Michele_Laino

question #1
for example if I have this rational function:
\[f\left( x \right) = \frac{1}{{{x^2} + 2x}}\]
then, in order to find the asymptotes, I try to factorize the quantity:
\(x^2+2x\)

- anonymous

0?

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## More answers

- Michele_Laino

hint: I try to factorize the denominator, since \(x^2+2x\) is the denominator of my function above. So what is the right option?

- anonymous

So C

- Michele_Laino

please wait a moment, what is the maning of "holes" of a function?

- anonymous

well its asking if theres any holes in the graph I'm not sure what it means

- Michele_Laino

yes! oops.. meaning*

- anonymous

How do we figure out number 2?

- Michele_Laino

I understand, the hole of a graph is a point x such that the function assumes an infinite value
so the answer to first question is option C, then you are right!

- Michele_Laino

I'm sorry, the right answer to question #1 is option D

- Michele_Laino

we have to factorize both numerator and denominator

- anonymous

okay what about the 2nd one?

- Michele_Laino

question #2
here we can factorize both numerator an denominator as follows:
\[f\left( x \right) = - \frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\]

- anonymous

i dont get what the answer choice would be?

- Michele_Laino

as I wrote before, in order to understand if the function has some holes, we have to factorize both numerator and denominator. Now, in my factorization above, you can cancel two factors, do you know what are such factor?

- anonymous

C?

- anonymous

or D could be the answers

- Michele_Laino

hint:
after a simplification, I can write this:
\[f\left( x \right) = - \frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = - \frac{{x + 3}}{{x + 2}}\]
as we can see at \(x=2\) our function is continuous, so, it can not be option C, and for the same reason it can not be option D

- anonymous

B

- Michele_Laino

that's right!

- anonymous

(: what about the 3rd one?

- Michele_Laino

question #3
hint:
such function is a rational function, and it is given by the quotient between these two polynomials:
\(x+2\) and \(x-2\)

- anonymous

A

- Michele_Laino

that's right!

- anonymous

Can you help me with 2 more ?

##### 1 Attachment

- Michele_Laino

question #4
as I wrote before, I can rewrite the function like below:
\[f\left( x \right) = - \frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = - \frac{{x + 3}}{{x + 2}}\]
now, please what are the values \(x\) such that the denominator is equal to zero?

- anonymous

2?

- Michele_Laino

we have to solve this equation:
\(x+2=0\), so what is \(x\) ?

- anonymous

-2

- Michele_Laino

that's right, so how many asymptotes we have?

- anonymous

0?

- Michele_Laino

please we have the only asymptote \(x=-2\), am I right?

- anonymous

D?

- Michele_Laino

\(x=-2\) is a vertical asymptote

- anonymous

oh so B then?

- Michele_Laino

that's right!

- anonymous

and #5?

- Michele_Laino

question #5
I have already answered to that question, when I wrote this formula:
\[\begin{gathered}
f\left( x \right) = \frac{{3 - 2x - {x^2}}}{{{x^2} + x - 2}} = \hfill \\
\hfill \\
= - \frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = - \frac{{x + 3}}{{x + 2}} \hfill \\
\end{gathered} \]

- anonymous

B

- Michele_Laino

that's right!
please, when you have to do similar exercises, remember to factorize both numerator and denominator of the rational function

- anonymous

Okay thank you so much! (:

- Michele_Laino

:)

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