Graphing f(x)=2^2+9/5x^2+2 with imaginary asymptotes

[Rasine]

i need to graph f(x)=2^2+9/5x^2+2

the vertical asym. would be the i squareroot10/2 ?

if that is correct how do i graph an imaginary #

 

[ToxicBug]

You're telling me you need to graph

$$9/5x^2+6$$

?

 

[Rasine]

no in the numerator is: 2x^2+9
denomorator is: 5x^2+2

 

[ToxicBug]

So it is

$$\frac{2x^2+9}{5x^2+2}$$

and you have to sketch the graph?

 

[ToxicBug]

It has horizontal asymptote at y=2/5 and no vertical asymptotes since the denominator is never zero.

 

[Rasine]

thank you very much

 

[dextercioby]

Yes,asymptotes (especially vertical ones) refer to REAL values for numbers...

Daniel.