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 $f(x).$
+33615 "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.)
Why would the horizontal asymptote mean that it it set to infinity and why would it be equal to a and -4? Could you please explain? Sorry, I really don’t understand...,.
+118608 Hi,
I struggled a little to get my head around it too. 
I'm letting y=f(x) some of the time because it is easier for me to work with.
there is a vertical asymptote at x=3 this means that y tends to =+ or - infinty when x=3
This will happen if the denominator tends to 0 as x tends to 3
hence c=-3
There is a horizontal asymptote at y=-4.
This means that as y tends to -4, x tends to =+ or - infinty
So
\(\displaystyle \lim_{x\rightarrow \infty}\; f(x)=-4\qquad or \qquad \displaystyle \lim_{x\rightarrow -\infty}\; f(x)=-4\\ a=-4\qquad or \qquad a=-4\\ so\;\;a=-4 \)
so we have
\(f(x) = \frac{-4x+b}{x-3}\)
Now sub in ( 1,0) to get the value of b
you will get b=4
\(f(x) = \frac{-4x+4}{x-3}\)
If you think about the asymptotes on a graph it might help. It did help me.
And here is the actual graph
