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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).

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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.

$$f(x)=\frac{rx+s}{2x+t}\\ y=\frac{rx+s}{2x-6}\\ y=\frac{rx+s}{2x-6}(2x-6)\\ y(2x-6)=rx+s\\ 2xy-4x=6y+s\\ 2x(y-2)=6y+s\\ 2x=\frac{6y+s}{y-2}\\ x=\frac{\frac{6y+s}{y-2}}{2}$$

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