Question

How do you graph `f(x)=(x^2-12x)/(x^2-2x-3)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

By considering asymptotes, yields;
graph{y = (x^2-12x)/(x^2-2x-3) [-19.82, 19.83, -9.9, 9.92]}

Explanation:

First we can factorise to give `f(x)` = `(x(x-12)) / ((x-3)(x+1))`

Now we can consider verticle asymptotes, hence at `x = 3 and x = -1`, Where the function is undefined, when the denominator `=0`

Now we can cosnider Horizontal asymptotes:

`lim_(x->+//-oo) f(x)` = `lim_(x->+//-oo) (1-12/x)/(1-2/x -3/(x^2))`

Hence `lim_(x->+//-oo) f(x)` = 1
as `a/oo -> 0`

Hence Horizontal asymptote ; `y = 1`

We now can consider roots; `f(x) = 0`

Hence `f(x) = 0 -> x(x-12) = 0`

Hence has roots, `x = 0 and x = 12`

Now can cosnider the nature of function as they approach the verticle asymptotes from possitive and negative direction:

We can compute the follwoing limits, via letting `x` be close to the limit value `x`, e.g. `lim_(x->3^-) f(x) ~ f(2.9999)` to give us a general idea to weather it tends to pos or neg `oo`

`lim_(x->3^-) f(x) = oo`
`lim_(x->3^+) f(x) = -oo`
`lim_(x->-1^-) f(x) = oo`
`lim_(x->-1^+) f(x) = -oo`

We this is efficiant information to be able to sketch this function,

So we note that as `x->3^+: y->-oo`, one of the roots is `x = 12`, and how as `x->oo : y ->1` yielding, graph{(x^2-12x)/(x^2-2x-3) [2.76, 23.975, -9.42, 1.19]}

We can repeat this for the other side of the asymptote, noting `f(0) = 0`

Hence;

graph{y = (x^2-12x)/(x^2-2x-3) [-19.82, 19.83, -9.9, 9.92]}