# A function f has horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).Part (a): Let f be of the form f(x) = {ax+b}/{x+c}. Find an expression for f(x). Part...

A function f has horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).Part (a): Let f be of the form f(x) = {ax+b}/{x+c}. Find an expression for f(x). Part (b): Let f be of the form f(x) = {rx+s}/{2x+t}. Find an expression for f(x).

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### 1 Answer

a) You need to determine the function` f(x) = (ax+b)/(x+c)` , hence, you need to calculate the coefficients a,b,c, under the conditions provided by the problem.

One condition is that the graph of function has an x intercept at (1,0), such that:

`f(1) = 0`

Replacing 1 for x, yields:

`(a + b)/(1 + c) = 0 => a + b = 0` , since 1 + c`!=` 0.

Other condition provided by the problem is that the function has a vertical asymptote of x = 3, hence, x - 3 = 0.

Since a function has a vertical asymptote of x = a for a being a value excluded from the domain of definition of function and since the function is a fraction and the values that cancel the denominator must be excluded from domain, yields that x + c = x - 3, hence c = -3.

Other condition provided by the problem is that the function has a horizontal asymptote of y =- 4, such that `lim_(x->+-oo) f(x) = -4.`

Evaluating the limit yields

`lim_(x->+-oo)(ax+b)/(x+c) = (+-oo)/(+-oo)`

Forcing the factor x yields:

`lim_(x->+-oo) (x(a + b/x))/(x(1 + c/x)) = lim_(x->+-oo) (a + b/x)/(1 + c/x)`

Since `lim_(x->+-oo) b/x = lim_(x->+-oo) c/x = 0` , yields:

`lim_(x->+-oo) (a + b/x)/(1 + c/x) = a/1 = a`

But `lim_(x->+-oo) f(x) = -4, hence a = -4.`

Since, from a previous condition yields a + b = 0, then b = 4.

**Hence, evaluating the function `f(x) = (ax+b)/(x+c),` under the given conditions, yields**` f(x) = (-4x+4)/(x-3).`

b) You need to determine the function `f(x) = (rx+s)/(2x+t)` , hence, you need to calculate the coefficients r,s,t, under the conditions provided by the problem.

One condition is that the graph of function has an x intercept at (1,0), such that:

`f(1) = 0`

Replacing 1 for x, yields:

`(r + s)/(2 + c) = 0 =>r + s = 0` , since `2 + c != 0.`

Other condition provided by the problem is that the function has a vertical asymptote of x = 3, hence, x - 3 = 0.

Since a function has a vertical asymptote of x = a for a being a value excluded from the domain of definition of function and since the function is a fraction and the values that cancel the denominator must be excluded from domain, yields that 2x + t = 0, for x = 3, hence t = -6.

Other condition provided by the problem is that the function has a horizontal asymptote of `y =- 4` , such that `lim_(x->+-oo) f(x) = -4.`

Evaluating the limit yields

`lim_(x->+-oo)(rx+s)/(2x+t) = (+-oo)/(+-oo)`

Forcing the factor x yields:

`lim_(x->+-oo) (x(r + s/x))/(x(2 + t/x)) = lim_(x->+-oo) (r + s/x)/(2 + t/x)`

Since `lim_(x->+-oo) s/x = lim_(x->+-oo) t/x = 0` , yields:

`lim_(x->+-oo)(r + s/x)/(2 + t/x) = r/2`

But `lim_(x->+-oo) f(x) = -4,` hence` r/2 = -4` , then r = -8.

Since, from a previous condition yields r`+s = 0` , then `s = 8.`

**Hence, evaluating the function `f(x) = (rx+s)/(2x+t)` , under the given conditions, yields `f(x) = (-8x+8)/(2x-6).` **