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A function \(f\) has a 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 \(f(x) = \frac{ax+b}{x+c}\)
Find an expression for \(f(x)\)

Part (b): Let \(f\) be of the form \(f(x) = \frac{rx+s}{2x+t}\)
Find an expression for \(f(x)\)

 Feb 11, 2017
 #1
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Part (a)

 

f(x) = (ax + b)/(x + c)

 

f(infinity) = - 4  so  - 4 = a       (limit as x goes to infinity of (ax+b)/(x+c) is a)   so a = -4

 

f(3) = infinite so  3 + c = 0     so c = -3

 

f(1) = 0        so  (a + b)/(1 + c) = 0   

                          (-4 + b)/(1 - 3) = 0      so b = 4

 

hence f(x) = (-4x + 4)/(x - 3)      or    f(x) = 4(1 - x)/(x - 3)

 

 

I'll leave you to do part (b).

 Feb 11, 2017

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