anonymous
  • anonymous
What are the points of discontinuity? Are they all removable? y = (x-3)/x^2 - 12x + 27
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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myininaya
  • myininaya
When is the bottom 0?
myininaya
  • myininaya
For what values of x make the bottom 0?
anonymous
  • anonymous
not sure?

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myininaya
  • myininaya
So you should be able to solve an equation in this form ax^2+bx+c=0 by now. I'm guessing this since this is a type of problem that occurs after learning that.
myininaya
  • myininaya
Do you know how to factor?
myininaya
  • myininaya
Or even use the quadratic formula to solve a quadratic equation?
myininaya
  • myininaya
You can even do completing the square.
anonymous
  • anonymous
so x^2+12x+27=0?
anonymous
  • anonymous
how do i do that?
myininaya
  • myininaya
Do you have -12x or +12x?
anonymous
  • anonymous
its postive 12x
myininaya
  • myininaya
Hmmm...So the problem above is written/typed wrongly I assume. Okay You should know how to factor the polynomial x^2+12x+27
myininaya
  • myininaya
Find two factors of c that add up to be b (you can do this when the coefficient of x^2 is 1 which in this problem it is)
myininaya
  • myininaya
In other words find two factors of 27 that add up to be 12.
anonymous
  • anonymous
its like this |dw:1370027288868:dw|
myininaya
  • myininaya
err so it is -12x?
anonymous
  • anonymous
noo x^2 minus 12x
myininaya
  • myininaya
I'm really confused what problem we are doing because twice now you have written +12x
myininaya
  • myininaya
You have also said twice that we have -12x
myininaya
  • myininaya
Anyways, we take the bottom polynomial whatever it may be and set that equal to zero and solve for x. I will let you do it since I don't know what problem we are doing anymore.
anonymous
  • anonymous
sorry
anonymous
  • anonymous
this is the problem were doing
1 Attachment
myininaya
  • myininaya
Set bottom polynomial equal to zero. \[x^2-12x+27=0\] You will notice this is in the form \[ax^2+bx+c=0\] You have learned if you cannot factor this easily you could also use completing the square or quadratic formula. The easy way to solve this quadratic is by factoring in my opinion. Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n. So The factored form would be (x-m)(x-n)=0 Then you set both factors equal to 0 like so x-m=0 or x-n=0 Solve both for x like so x=m or x=n
anonymous
  • anonymous
so which numbers do we use for x-m and x-n?
myininaya
  • myininaya
Have you found factors of 27 that have product 27 and sum -12?
myininaya
  • myininaya
Those are your m and n.
myininaya
  • myininaya
Recall this part that I said: Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.
myininaya
  • myininaya
c is 27 and b is -12
myininaya
  • myininaya
|dw:1370027980807:dw|
myininaya
  • myininaya
See how I factored 27?
myininaya
  • myininaya
1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27
anonymous
  • anonymous
Yes I see on my answes I have x= -9, x =-3 is that right?
myininaya
  • myininaya
Do any of those factors above that multiplied to be 27 also add up to be -12?
anonymous
  • anonymous
are these numbers removable?
anonymous
  • anonymous
and -3?
myininaya
  • myininaya
(x-3)(x-9)=0 Set both equal to 0.
myininaya
  • myininaya
I put (x-m)(x-n)=0 where I meant (x+m)(x+n)=0. So you would have x+m=0 or x+n=0 Which means you would have x=-m or x=-n. But anyways we have (x-3)(x-9)=0 And you are to set both factors equal to 0 and solve for x for both equations
myininaya
  • myininaya
x-3=0 or x-9=0 I leaving solving these linear equations to you.
anonymous
  • anonymous
do we also use -1 or no?
myininaya
  • myininaya
Where does that come from?
myininaya
  • myininaya
Have you solve x-3=0 or x-9=0 yet?
anonymous
  • anonymous
no idk how
myininaya
  • myininaya
x-3=0 You can't solve this for x? You don't know how to get x by itself?
myininaya
  • myininaya
To undo subtraction by three you add three.
myininaya
  • myininaya
Try adding three to both sides and tell me what the result is.
myininaya
  • myininaya
x-3=0 add 3 to both sides +3 +3 --------- what is the result?
myininaya
  • myininaya
x-3+3=0+3
myininaya
  • myininaya
what is -3+3? what is 0+3?
anonymous
  • anonymous
0 and 3
myininaya
  • myininaya
yes so x-3+3=0+3 gives us x+0=3 Or just x=3
myininaya
  • myininaya
Now you solve x-9=0
myininaya
  • myininaya
What are you going to add to both sides?
anonymous
  • anonymous
9?
myininaya
  • myininaya
yes so x=?
anonymous
  • anonymous
9 and 0?
anonymous
  • anonymous
so x=9
myininaya
  • myininaya
yes
myininaya
  • myininaya
Discontinuous at x=3 or x=9
myininaya
  • myininaya
Now go back to the problem where you had everything factored.
myininaya
  • myininaya
\[\frac{x-3}{(x-3)(x-9)}\]
myininaya
  • myininaya
This was the problem where everything is factored, correct?
myininaya
  • myininaya
Does anything cancel? If so that is where you have your removable discontinuity.
myininaya
  • myininaya
x-3 is just a number so is x-9 If we had 5/(5*9) you would say the 5's cancel right? Like this 1/9 Does anything cancel? Is there a factor on top that you see on bottom?
myininaya
  • myininaya
\[\frac{(x-3)}{(x-3)(x-9)}\] If you want you can put parenthesis around the x-3 on top, Do you see any factors that match on top and bottom?
myininaya
  • myininaya
This is just like matching socks.
myininaya
  • myininaya
I have to go to the grocery store. Good luck.
anonymous
  • anonymous
Thanks for all ur help

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