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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).

 

a) Let f be of the form \(f(x) = \frac{rx+s}{2x+t}\). Find an expression for f(x).

 

b) Let f be of the form \(f(x) = \frac{ax+b}{x+c}\). Find an expression for f(x).

 Jun 24, 2020
 #1
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(a) We use the vertical asymptote, x=3, to help us solve for c. We let x be 3. We need to find c so that the denominator, x+c, is 0, and the fraction is undefined.

\(x+c=0\\3+c=0\\c=-3.\) 

 

We let f(x) equal to y and plug in the value of c. We then have

\(y=\frac{ax+b}{x-3}\\y(x-3)=\frac{ax+b}{x-3}(x-3)\\y(x-3)=ax+b\\xy-3y=ax+b\\xy-ax=b+3y\\x(y-a)=b+3y\\x=\frac{3y+b}{-a+y}.\) 

Now we use the horizontal asymptote y=-4 to help us solve for a. Using similar logic as we solved for c to solve for a, we get

 \(-a+y=0\\-a-4=0\\a=-4.\)

 

We then plug the values of a and c into \(f(x)=\frac{ax+b}{x-3}\) to get \(f(x)=\frac{-4x+b}{x-3}\). We plug in the point (1, 0) into the equation to solve for b. \(f(x)=\frac{-4x+b}{x-3}\\0=\frac{-4(1)+b}{1-3}\\0=\frac{-4+b}{-2}\\0=-4+b\\b=4.\)

 

Therefore \(f(x) = \frac{ax+b}{x+c}\) is \(\boxed{f(x) = \frac{-4x+4}{x-3}.}\) 

 

Use similar logic to solve for part (b).

 Jun 24, 2020
 #2
avatar+313 
+3

Thank you so much!!

mathmathj28  Jun 25, 2020
 #3
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I'm a bit confused. Why do we want the fraction to be undefined? Could you please explain this to me?

Thanks

Guest Jun 28, 2020

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