Sketch the graph of this function.

Question

abbles · Answer

(2x^3 - 1)/(5x^3 - 16)

abbles · Answer

abbles · Answer

The horizontal asymptote is 2/5 but I don't know how to find the vertical ones.

sweetburger · Answer

basically you just set the denominator equal to 0 and solve for x
$$5x^3-16=0$$

anthonyym · Answer

This is because if the denominator equals 0, the fraction would be undefined (you can't divide by 0).

abbles · Answer

Thanks!  So the vertical asymptote would be the cube root of 16/5?

sweetburger · Answer

yep

abbles · Answer

Now i now the vertical asymptote and horiz. one... what next?  plug in a few points?

abbles · Answer

Is the vertical asymptote 0.5 or 1.5?  Plugging in some points, it seems to be at 1.5 but the cube root of 16/5 is 0.5...

abbles · Answer

Nevermind I think I got this

sweetburger · Answer

I will link a desmos graph real quick

sweetburger · Answer

https://gyazo.com/361967ec36cfe42ccb82db8109d5ca22 also if you take $$\sqrt[3]{\frac{ 16 }{ 5 }}$$ you will get roughly 1.5

abbles · Answer

Yeah, I forgot to put parenthesis around 16/5 when I took the cube root.  Thanks for the graph!  It looks pretty much like the mine :)

sweetburger · Answer

Ahh ok that makes sense why the point was coming to .5 . Alright glad you sorted it out :)