Help with V.A. and H.A. and domain

[janetallen06]

Hi, this is my first post. The equation is 2s^2/3s^2-5. I got 2/3 as the H.A. but am stuck on V.A. and domain. Any help would be great. Thank you

 

[galactus]

The vertical asymptote occurs where the denominator equals 0.

Solve \(\displaystyle 3s^{2}-5=0\) for s to find the VA's.

The domain is all the values which will provide real solutions. Since division by 0 is undefined, those values will not be in the domain.

 

[janetallen06]

When I solve for s i end up with a square root answer, still confused

 

[stapel]

The equation is 2s^2/3s^2-5.

Do you mean the "expression" (since there is no "equals" sign), or is this an equation with a name, such as "y" or "f(x)"?

Do you mean "(2s[sup:2x9pfyze]2[/sup:2x9pfyze]/3s[sup:2x9pfyze]2[/sup:2x9pfyze]) - 5", as posted, or did you mean (as the previous reply assumed) "(2s[sup:2x9pfyze]2[/sup:2x9pfyze]) / (3s[sup:2x9pfyze]2[/sup:2x9pfyze] - 5)"?

I got 2/3 as the H.A. but am stuck on V.A. and domain.

Was the previous assumption correct, that you mean "horizontal asymptote" and "vertical asymptote"?

When I solve for s i end up with a square root answer, still confused

Has your class not yet covered how to solve quadratic equations?

Thank you!

Eliz.

 

[janetallen06]

Re:

The equation is 2s^2/3s^2-5.

Do you mean the "expression" (since there is no "equals" sign), or is this an equation with a name, such as "y" or "f(x)"?

Do you mean "(2s[sup:1nzh7gkk]2[/sup:1nzh7gkk]/3s[sup:1nzh7gkk]2[/sup:1nzh7gkk]) - 5", as posted, or did you mean (as the previous reply assumed) "(2s[sup:1nzh7gkk]2[/sup:1nzh7gkk]) / (3s[sup:1nzh7gkk]2[/sup:1nzh7gkk] - 5)"?

I got 2/3 as the H.A. but am stuck on V.A. and domain.

Was the previous assumption correct, that you mean "horizontal asymptote" and "vertical asymptote"?

When I solve for s i end up with a square root answer, still confused

Has your class not yet covered how to solve quadratic equations?

Thank you!

Eliz.

Yes we have covered solving quadratic equations, but my mother passed away on Monday and I missed all week and they finished solving quadratic equations and Vertical and Horizontal Asymptotes. I am trying to learn this on my own. Yes, I should have set it to f(x)=

 

[galactus]

Sorry to hear about your mother. Must be rough concentrating on math right now.

You can have a solution with a radical. It does not have to be a nice integer solution. What did you get?.

 

[janetallen06]

Sorry to hear about your mother. Must be rough concentrating on math right now.

You can have a solution with a radical. It does not have to be a nice integer solution. What did you get?.

i got the square root of 5/3

 

[galactus]

That's it. Except there are two of them. \(\displaystyle \pm\sqrt{\frac{5}{3}}\)

Which can be written as \(\displaystyle \pm\frac{\sqrt{15}}{3}\)

The domain is all values except these.