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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 $f(x).$

 Sep 2, 2017

Best Answer 

 #1
avatar+8044 
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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 $f(x).$

 

Part (a) 

\(\large f(x)=\frac{8}{3-x}-4=\frac{8-12+4x}{3-x}\\ f(x) = \frac{ax+b}{x+c}\\ \large \color{blue}f(x)=\frac{-4x+4}{x-3}\) 

 

a = -4; b = 4; c = -3

 

 

Part (b)

\(f(x)=\frac{rx+s}{2x+t}\\ \large \color{blue}f(x)=\frac{-8x+8}{2x-6}\) 

 

r = -8; s = 8; t = -6

 

laugh  !

 Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017
 #1
avatar+8044 
+1
Best Answer

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 $f(x).$

 

Part (a) 

\(\large f(x)=\frac{8}{3-x}-4=\frac{8-12+4x}{3-x}\\ f(x) = \frac{ax+b}{x+c}\\ \large \color{blue}f(x)=\frac{-4x+4}{x-3}\) 

 

a = -4; b = 4; c = -3

 

 

Part (b)

\(f(x)=\frac{rx+s}{2x+t}\\ \large \color{blue}f(x)=\frac{-8x+8}{2x-6}\) 

 

r = -8; s = 8; t = -6

 

laugh  !

asinus Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017
 #2
avatar+8044 
+1

The graph

 

\(f(x)=\frac{rx+s}{2x+t}\\ \large \color{blue}f(x)=\frac{-8x+8}{2x-6}\)

 

laugh  !

 Sep 3, 2017

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