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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
0

(a) f(x) = (3x + 2)/(x + 7)

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

Apr 20, 2020
#2
+9447
+1

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

(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

!

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

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

(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)

!

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