Question

How do you graph `f(x)=(2x-1)/(x^3-9x)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

x intercept:
`A=(1/2,0)`

y intercept:
`NaN`

vertical asymptotes:
`x=0`
`x=3`
`x=-3`

horizontal asymptotes:
`y=0` (for both x very big positive and very big negative)

Explanation:

FIRST thing we ALWAYS do is making the function easier:

`f_((x))=(2x-1)/((x^3-9x)`
`=> (2x-1)/(x(x^2-9))`
`=> (2x-1)/(x(x^2-9))`
`=> (2x-1)/(x(x-3)(x+3)` because `(a^2-b^2)=(a-b)(a+b)`

Now we can "easily" see that `x` MUST NOT be 0,3,-3 because we know that we must not divided by 0.
We can also see that `(2x-1=0)` if `(x=1/2)`

vertical asymptotes:

Therefore we now know that the function has 3 vertical asymptotes (we get them when `x=0` in the "down side of the equation"):
`x=0`
`x=3`
`x=-3`

horizontal asymptotes:
(I don't know if you know how to use `lim` so I will explain in an alternative way)

horizontal asymptotes is a question of a kind of "What does the function "act" when `x` is REALLY HUGE (and/or negative HUGE)?

To solve it easily, lets check with a calculator what happens when x comes is really big, for example I will check the positives:

`f_((1,000,000))=1.999999*10^-12=0.0000000000000001999`
`f_((1,00,000,000))=1.9999*10^-15=0.0000000000000000001999`

So, the function goes to `y=0` for very big x-es.
On the same way, we can chek the very negative x-es:
`f_((1,000,000))=2.000001*10^-12=0.00000000000000020001`

So, in that case, the function goes to `y=0` for very negative big x-es.

therefore, the horizontal asymptote is:
`y=0`

x and y intercepts:
Now that we have an "easier" function:
`(f_((x))=(2x-1)/(x(x-3)(x+3)))`

We check both "x=0" and y=0":
`f_((0))=(2*0-1)/(0*(0-3)(0+3))=-1/0=NaN`
But we already knew it because `x=0` is one of ours' vertical asymptotes...
Now lets check what if `f_((x))=0` what is x?
`(2*x-1)/(x*(x-3)(x+3))=0`
`=> (2*x-1)=0` (for x in not 1,3,-3)
`=> 2x=1`
`=> x=1/2`

So we have only one point `A=(1/2,0)`