Question

How do you write an equation with a vertical asymptote of `3`, slant asymptote of `y=x+1`, and `x` intercept at `2`?

Answer 1

`f(x) = (x^2-2x)/(x-3)`

Explanation:

Let `f(x) = (n(x))/(d(x)` be the choosed fractional function.

It has a vertical asymptote at `3` so `d(x) = (x-3)`
It has a slant symptote given by `y = x+1` so `n(x)` degree must be `2`.

Now, the general polynomial of degree `2` is

`n(x) = a x^2+b x+ c` so

`f(x) = (a x^2+b x+ c)/(x-3)`

We know also that `f(2) = 0` and

`a x^2+b x+ c=(x-3) (x+1) + d` because

`f(x) = (a x^2+b x+ c)/(x-3) = (x+1)+d/(x-3)` so for big values of `abs x`

`f(x) approx (x+1)`

then

`a x^2+b x+ c = x^2-2x+d-3`

so `a = 1, b = -2, c = d-3` but from `f(2) = 0` we extract

`0 = 4a+2b+c = 4-4+c->c=0->d=3`

Finally

`f(x) = (x^2-2x)/(x-3)`