"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). "
\(f(x)=\frac{ax+b}{x+c}\)
Horizontal asymptote means set x to infinity so \(f(x)\rightarrow a\) hence we immediately have \(a=-4\)
Vertical asymptote means set denominator to zero at x = 3, so \(3+c=0 \text{ or } c = -3\)
x intercept at (1, 0) means \(\frac{a\times1+b}{1+c}=0\) so \(\frac{-4+b}{1-3}=0 \text{ or }b = 4\)
Hence \(f(x)=\frac{-4x+4}{x-3}\text{ or }f(x)=\frac{4(1-x)}{x-3}\)
Use the same approach for part b.
(Edited to correct silly mistake! Thanks heureka.)
.