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What are the points of discontinuity? Are they all removable?
y = (x3)/x^2  12x + 27
 10 months ago
 10 months ago
What are the points of discontinuity? Are they all removable? y = (x3)/x^2  12x + 27
 10 months ago
 10 months ago

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myininayaBest ResponseYou've already chosen the best response.1
When is the bottom 0?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
For what values of x make the bottom 0?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
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.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Do you know how to factor?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Or even use the quadratic formula to solve a quadratic equation?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
You can even do completing the square.
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
so x^2+12x+27=0?
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
how do i do that?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Do you have 12x or +12x?
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
its postive 12x
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Hmmm...So the problem above is written/typed wrongly I assume. Okay You should know how to factor the polynomial x^2+12x+27
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
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)
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
In other words find two factors of 27 that add up to be 12.
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
its like this dw:1370027288868:dw
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
noo x^2 minus 12x
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
I'm really confused what problem we are doing because twice now you have written +12x
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
You have also said twice that we have 12x
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
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.
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
this is the problem were doing
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Set bottom polynomial equal to zero. \[x^212x+27=0\] 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 (xm)(xn)=0 Then you set both factors equal to 0 like so xm=0 or xn=0 Solve both for x like so x=m or x=n
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
so which numbers do we use for xm and xn?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Have you found factors of 27 that have product 27 and sum 12?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Those are your m and n.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
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.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
c is 27 and b is 12
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
dw:1370027980807:dw
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
See how I factored 27?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
1(27)=27 3(9)=27 1(27)=27 3(9)=27
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
Yes I see on my answes I have x= 9, x =3 is that right?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Do any of those factors above that multiplied to be 27 also add up to be 12?
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
are these numbers removable?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
(x3)(x9)=0 Set both equal to 0.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
I put (xm)(xn)=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 (x3)(x9)=0 And you are to set both factors equal to 0 and solve for x for both equations
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
x3=0 or x9=0 I leaving solving these linear equations to you.
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
do we also use 1 or no?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Where does that come from?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Have you solve x3=0 or x9=0 yet?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
x3=0 You can't solve this for x? You don't know how to get x by itself?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
To undo subtraction by three you add three.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Try adding three to both sides and tell me what the result is.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
x3=0 add 3 to both sides +3 +3  what is the result?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
what is 3+3? what is 0+3?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
yes so x3+3=0+3 gives us x+0=3 Or just x=3
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Now you solve x9=0
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
What are you going to add to both sides?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Discontinuous at x=3 or x=9
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Now go back to the problem where you had everything factored.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
\[\frac{x3}{(x3)(x9)}\]
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
This was the problem where everything is factored, correct?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Does anything cancel? If so that is where you have your removable discontinuity.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
x3 is just a number so is x9 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?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
\[\frac{(x3)}{(x3)(x9)}\] If you want you can put parenthesis around the x3 on top, Do you see any factors that match on top and bottom?
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
This is just like matching socks.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
I have to go to the grocery store. Good luck.
 10 months ago

algebra2sucksBest ResponseYou've already chosen the best response.0
Thanks for all ur help
 10 months ago
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