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 $f(x)$

Guest Sep 3, 2018

edited by
Guest
Sep 3, 2018

#1**+1 **

\(f(x) = \dfrac{a x + b}{x + c} \\ \\ \text{there is a vertical asymptote at }x=3 \text{ so clearly c=-3} \\ \\ \lim \limits_{x \to \pm \infty} f(x) = a \text{ so }a=-4 \\ \\ f(x) = \dfrac{-4x + b}{x-3} \\ \\ f(1) = \dfrac{-4+b}{1-3} = 0 \\ \\ \text{so clearly b=4} \text{ and thus }f(x) = \dfrac{-4x+4}{x-3}\)

see if you can work out part (b) using this as a template

Guest Sep 4, 2018