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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).

 

Thanks in advance!

 Aug 21, 2018
 #1
avatar+98197 
+1

A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).   Find an expression for f(x)....

 

a)   ax + b

     ______  =  f(x)       

       x + c

 

If the vertical asymptote is 3, then , in the denominator,  3 + c  = 0  and  c   =  -3

If the hrizontla asymptote is -4, then  ax / x  = -4  ⇒  a   = -4

And if the x intercept = ( 1,0), this implies that   a(1) + b  = 0  ⇒  -4(1) + b  = 0  ⇒

-4 + b  = 0  ⇒   b  = 4

 

Here is a graph : https://www.desmos.com/calculator/ohx3qy9kvt

 

 

cool cool cool

 Aug 21, 2018
 #2
avatar+98197 
+1

b )     rx + s

        _____   =  f(x)

        2x + t

 

If the horizontal asymptote  =  -4, then   r / 2  = -4 ⇒  r  = -8

If the vertical asymptote = 3, then  2(3) + t  = 0  ⇒  6 + t  = 0  ⇒  t   = -6

If  (1,0)  is an x intercept, then  r(1) + s  = 0 ⇒  -8(1) + s  = 0 ⇒ - 8 + s  = 0 ⇒

s  = 8

 

Here's a graph :  https://www.desmos.com/calculator/cown9bo626

 

 

cool cool cool

 Aug 21, 2018
 #3
avatar+973 
+1

Thanks Chris!

Lightning  Aug 21, 2018
 #4
avatar+98197 
0

OK, Lightning....glad I could help   !!!

 

cool cool cool

CPhill  Aug 21, 2018

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