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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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).`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial