OpenStudy (anonymous):

can someone please solve these for me :( 1. y = 2x^2 + 5 over x^2 2. y= 3x - 3 over x^2 - 1 3. y = x^2 + 5x+ 6 over x^2 + 6x + 9 4. y =x ^3 -8 over x^2 -8 5. y= 1 over 2x^2 + 3x - 7 please help me!

5 years ago
OpenStudy (amistre64):

what does it mean to "solve"?

5 years ago
OpenStudy (anonymous):

the question says " find any points of discontinuity for each rational function"

5 years ago
OpenStudy (anonymous):

@amistre64

5 years ago
OpenStudy (amistre64):

discontinuity in this case usually defines a value for which the function becomes undefined at. The values for when the denominators go to zero

5 years ago
OpenStudy (anonymous):

oh okay , so could u plz answer them for me

5 years ago
OpenStudy (amistre64):

answer them for you is a different website. We promote studying and learning of the material over handing out specifics. I would stay and see if i could guide you thru them, but i have class starting pretty soon

5 years ago
OpenStudy (amistre64):

3 and 5 seem to be about the trickest ones to solve for; the others are pretty straight forward

5 years ago
OpenStudy (anonymous):

oh okay , so could u jus help me with the first problem ? like how do i solve it ?

5 years ago
OpenStudy (amistre64):

y = 2x^2 + 5 over x^2 when does x^2 = 0 ?

5 years ago
OpenStudy (amistre64):

y= 3x - 3 over x^2 - 1 when does x^2 - 1 = 0 ?

5 years ago
OpenStudy (anonymous):

um .... i dont know

5 years ago
OpenStudy (anonymous):

\[\frac{3x-3}{x^2-1}\] When \[{x^2-1}=0\], the function breaks and there is a discontinuity (you can't divide by 0)> \[x^2-1=0\] \[x^2=1\] Do you know what to do from here?

5 years ago
OpenStudy (anonymous):

no i dont know any of this at all

5 years ago
OpenStudy (anonymous):

Are you familiar with the fact that you can't divide by 0?

5 years ago
OpenStudy (anonymous):

yea i know that fact

5 years ago
OpenStudy (anonymous):

A discontinuity is where the function breaks down (that is, the input doesn't make sense when you perform the function on it). For example, the function 1/x breaks when you insert x=0, as you can't divide by 0. Same thing here.

5 years ago
OpenStudy (anonymous):

oh okay so what would be the answer for the first question ?

5 years ago
OpenStudy (anonymous):

Which x do you put into the function that makes there be a 0 in the denominator?

5 years ago
OpenStudy (anonymous):

um ... 2x?

5 years ago
OpenStudy (anonymous):

What number?|dw:1351705215913:dw|

5 years ago
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