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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 $f(x).$

 Nov 19, 2019
 #1
avatar+106515 
+1

If the first function has a horizontal asymptote at  y  =-4

Then the ratio of the coefficients on x in the numerator and denominator is   a / 1  = -4

So a  = -4

 

If it has an x itercept at (1,0)....then   

-4(1) +  b  = 0

-4 + b  = 0

b = 4

 

And if it has a vertical asymptote at 3.....then

3 + c   =0

c = -3

 

So   the function is  

 

             -4x + 4

f(x)  =    _____

               x - 3

 

Here is a graph :    https://www.desmos.com/calculator/pe4jejs6zo

 

 

cool cool cool

 Nov 19, 2019
 #2
avatar+106515 
+1

The second one uses similar reasoning

 

r / 2  = -4

r = -8

 

-8 ( 1)  + s  = 0

-8 + s  = 0

s = 8

 

2(3) + t  = 0

6 + t  = 0

t = -6

 

 

So

 

               -8x  +  8     

f(x)  =    ________

                2x  -   6

 

Here is the graph : https://www.desmos.com/calculator/oo1vhoudky

 

 

cool cool cool

 Nov 19, 2019

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