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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

 

ax+b/x+c

 

Find an expression for f(x)

Part (b): Let  be of the form

 

rx+s/2x+t


Find an expression for f(x)

 Sep 15, 2019
 #1
avatar+111396 
+1

(a)     ax + b

       ______

         x + c

 

If the vetical asymptote  =  3    then we want to solve this :  3 + c  = 0   →   c   = -3

 

If the horizontal asymptote  = - 4    then   a/1  = -4  →  a   =-4

 

Lastly  ......if the x intercept is (1, 0)  then  we need to solve this : 

 

-4(1) + b  = 0  →  -4 + b = 0 →  b  = 4

 

So  f(x)   =          -4x + 4

                       _________

                            x  - 3

 

 

cool cool cool

 Sep 15, 2019
 #2
avatar+111396 
+1

b)     rx + s

       ______

        2x + t

 

Using the same info

If the vertical asymptote  = 3....we need to solve this for t :   

 2(3) + t  = 0 →  6 + t  = 0 → t = -6

 

If the horizontal asymptote = -4, the we need to solve this for r :

r/2  = -4      multiply both sides by 2

r  = -8

 

And since (1,0)  is on the graph,  then we can solve this for s :

-8 (1)  + s  = 0

-8 + s  = 0

s = 8

 

So

 

f(x)   =      -8x + 8

               _______

                 2x - 6

 

 

cool cool cool

 Sep 15, 2019

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