Question

Im confused can someone help me?

Answer 1

@Michele_Laino i dont understand how to solve these?

Answer 2

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:
\(x^2+2x\)

Answer 3

0?

Answer 4

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

Answer 5

So C

Answer 6

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

Answer 7

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

Answer 8

yes! oops.. meaning*

Answer 9

How do we figure out number 2?

Answer 10

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!

Answer 11

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

Answer 12

we have to factorize both numerator and denominator

Answer 13

okay what about the 2nd one?

Answer 14

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

Answer 15

i dont get what the answer choice would be?

Answer 16

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?

Answer 17

C?

Answer 18

or D could be the answers

Answer 19

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

Answer 20

B

Answer 21

that's right!

Answer 22

(: what about the 3rd one?

Answer 23

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

Answer 24

A

Answer 25

that's right!

Answer 26

Can you help me with 2 more ?

Answer 27

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?

Answer 28

2?

Answer 29

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

Answer 30

-2

Answer 31

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

Answer 32

0?

Answer 33

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

Answer 34

D?

Answer 35

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

Answer 36

oh so B then?

Answer 37

that's right!

Answer 38

and #5?

Answer 39

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

Answer 40

B

Answer 41

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

Answer 42

Okay thank you so much! (:

Answer 43

:)