A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).
Let f be of the form \(f(x) = \frac{rx+s}{2x+t}\). Find an expression for f(x).
So what I tried doing is:
Using the vertical asymptote x=3, we let x=3 in the denominator. The fraction has to be undefined, so the denominator must be 0, therefore t= -6. Then I let f(x) equal to y.
\(f(x)=\frac{rx+s}{2x+t}\\ y=\frac{rx+s}{2x-6}\\ y=\frac{rx+s}{2x-6}(2x-6)\\ y(2x-6)=rx+s\\ 2xy-4x=6y+s\\ 2x(y-2)=6y+s\\ 2x=\frac{6y+s}{y-2}\\ x=\frac{\frac{6y+s}{y-2}}{2}\)
Then I don't know where to go from there? Someone please help me.
+310 I'm not sure how you went from \(y(2x-6)=rx+s\) to \(2xy-4x=6y+s\).
You expanded this part wrong, so let's start from this point.
\(y(2x-6)=rx+s\)
\(2xy-6y=rx+s\)
\(2xy-rx=6y+s\)
\(x(2y-r)=6y+s\)
\(x=\frac{6y+s}{2y-r}\)
Using the horizontal asymptote of y = -4, we find that r = -8.
Now use the point (1, 0) to find s.
