# a rational function that has the following characteristics: crosses the x-axis at 3; the x-axis at -2; one vert. asymp., x = 1; and one horiz. asymp.,y+2

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You need to consider the rational function as `f(x) = (p(x))/(q(x)).`

The problem provides the information that the graph of the function crosses `x` axis at `3` , hence, if `y = 0, x = 3` such that:

`(p(3))/(q(3)) = 0 => {(p(3) = 0),(q(3)!=0):}` ( `x = 3` is the root of p(x) but it cannot cancel the denominator q(x))

The problem provides the information that the graph of the function crosses `x` axis at `-2,` hence, if `y = 0, x =-2` such that:

`(p(-2))/(q(-2)) = 0 => {(p(-2) = 0),(q(-2)!=0):}` ( `x =-2` is the root of p(x) but it cannot cancel the denominator q(x))

The problem provides the information that the function has one vertical asymptote at `x = 1` , hence `lim_(x->1) f(x) = +-oo => x = 1` is the root of denominator, such that`q(1) = 0` .

Notice that the numerator has two roots and denominator has one root, hence, the function may be `f(x) = (a(x - 3)(x + 2))/(x - 1)`

The problem also provides the information that the function has an horizontal asymptote at `y = 2` , such that:

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

`lim_(x->+-oo) (ax^2 - ax - 6a)/(x - 1) = 2`

Since substituting `oo` for `x` in limit yields `oo/oo` , you may use l'Hospital's theorem such that:

`lim_(x->+-oo) (2ax - a)/1 = 2`

Since, substituting `oo` for `x` yields `oo` and not 2, then the information that the function would have an horizontal asymptote at `y = 2` is not valid, as the graph below proves it.

**Hence, evaluating the rational function, under the given conditions, yields, for `a = 1` , `f(x) = ((x - 3)(x + 2))/(x - 1).` **