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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). 

 

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).

 

I don't really know what to do. I formed the equations \(f(1)=\frac{x+b}{1+c}\) and \(f(1)=\frac{x+s}{2+t} \), respectively getting that \(\frac{x+b}{1+c}=0\) and \(\frac{x+s}{2+t}=0\). Then should I set them to equal each other? I don't really know if that would help because I'd still have a bunch of variables...am I even using the right approach? Also, I don't know if I should be utilizing the asymptotes to help me solve this.

 Jun 14, 2020
 #1
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(a) By definition of horizontal and vertical asymptotes:

 

\(\begin{cases} \displaystyle\lim_{x\to\infty} f(x) = -4\\ 3 + c = 0\\ \dfrac{a + b}{1 + c} = 0 \end{cases}\)

That means 

\(\begin{cases} a = -4\\ c = -3\\ b = 4 \end{cases}\)

(a, b, c) = (-4, 4, -3).

 

(b) is similar to (a).

 Jun 14, 2020
 #2
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I don't really know calculus yet so I don't really understand this. Is there another way?

 Jun 14, 2020
 #3
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The definition of a horizontal asymptote is a horizontal line to which the graph becomes closer and closer as x becomes a very large or very small number, but never actually reaches it. So it's a y value that's undefined. A vertical asymptote is the same thing, but an undefined value for x. 

 

a) Since the vertical asymptote is x=3, we can see that c = -3, since placing x=3 into \(\frac{ax+b}{x-3}\) is undefined. (The denominator of a fraction is undefined for 0). 

 

Let f(x) equal y. Now we can find a by solving this equation for x.

\(x=\frac{3y-b}{a-y}\). Using the same thought process as above, we find that a = -4

 

Now we can use our values c = -3 and a = -4 and plug them in:

\(y=\frac{-4x+b}{x-3}\)

 

All we have to do is find b. We still have on piece of information we haven't used: the point (1, 0). Plug these x and y values in. 

\(0=\frac{-4+b}{1-3}\)

Now we can solve for b and find that b = 4. 

So putting everything together, we have our final answer: \(f(x)=\frac{-4x+4}{x-3}\)

 

You can do the same thing for part b.

 Jun 14, 2020
 #4
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Oh thanks, this explanation helps a lot more.

 Jun 14, 2020
 #5
avatar+307 
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No problem! Glad it helped :)

thelizzybeth  Jun 14, 2020
 #6
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Wait, why is it \(x=\frac{3y-b}{a-y}\)?

Guest Jun 15, 2020

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