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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

#1
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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 !

Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017
edited by asinus  Sep 3, 2017

#1
+1

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 !

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

The graph

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

Sep 3, 2017