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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) = (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).

 Apr 20, 2020
edited by daddypig  Apr 20, 2020
edited by daddypig  Apr 20, 2020
edited by daddypig  Apr 20, 2020
 #1
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(a) f(x) = (3x + 2)/(x + 7)

 

(b) f(x) = (4x + 1)/(2x - 3)

 Apr 20, 2020
 #2
avatar+9447 
+1

(a): Let f be of the form f(x) = (ax+b)/(x+c).Find an expression for f(x).

 

Hello daddypig!

 

(a)

\(\color{blue}f(x)=\frac{1}{x-3}-4\\ f(x)=\frac{1}{x-3}-\frac{4(x-3)}{x-3}\\ f(x)=\frac{1-(4x-12)}{x-3}\)

 

\( f(x)=\frac{-4x+13}{x-3}\)   

 

\(\large f(x)=\frac{-4x+13}{x-3}\)  

 

a = - 4

b = 13

c = - 3

laugh  !

 Apr 20, 2020
edited by asinus  Apr 20, 2020
edited by asinus  Apr 20, 2020
 #3
avatar+9447 
+1

 (b): Let f be of the form f(x) = (rx+s)/(2x+t).Find an expression for f(x).

 

Hello daddypig!

 

(b)

 

\(\color{blue}f(x)=\frac{1}{x-3}-4\\ f(x)=\frac{1}{x-3}-\frac{4(x-3)}{x-3}\\ f(x)=\frac{1-(4x-12)}{x-3} \)

              \(2x+t=x-3\\ t=-x-3\)

 

\(\large f(x)=\frac{-4x+13}{2x+(-x-3)}\)

 

r = - 4

s = 13

t = (- x - 3)

laugh  !

 Apr 20, 2020
edited by asinus  Apr 20, 2020
edited by asinus  Apr 20, 2020

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