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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).

 

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.

 Jun 16, 2020
 #1
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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. 

 Jun 16, 2020
 #2
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Oh, oops! I see it now. I don't know where I got the -4x from. Thanks for helping me out!

 Jun 16, 2020

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