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# Help!!!

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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 $f(x).$

Aug 23, 2018

#1
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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). "

$$f(x)=\frac{ax+b}{x+c}$$

Horizontal asymptote means set x to infinity so $$f(x)\rightarrow a$$  hence we immediately have $$a=-4$$

Vertical asymptote means set denominator to zero at x = 3, so $$3+c=0 \text{ or } c = -3$$

x intercept at (1, 0) means $$\frac{a\times1+b}{1+c}=0$$  so  $$\frac{-4+b}{1-3}=0 \text{ or }b = 4$$

Hence $$f(x)=\frac{-4x+4}{x-3}\text{ or }f(x)=\frac{4(1-x)}{x-3}$$

Use the same approach for part b.

(Edited to correct silly mistake!  Thanks heureka.)

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Aug 23, 2018
edited by Alan  Aug 23, 2018
edited by Alan  Aug 23, 2018
#2
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Why would the horizontal asymptote mean that it it set to infinity and why would it be equal to a and -4? Could you please explain? Sorry, I really don’t understand...,.

Guest Aug 23, 2018
#3
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Hi,

I struggled a little to get my head around it too.

I'm letting y=f(x) some of the time because it is easier for me to work with.

there is a vertical asymptote at x=3 this means that  y tends to =+ or - infinty when x=3

This will happen if the denominator tends to 0 as x tends to 3

hence c=-3

There is a horizontal asymptote at y=-4.

This means that as y tends to  -4,     x  tends to =+ or - infinty

So

$$\displaystyle \lim_{x\rightarrow \infty}\; f(x)=-4\qquad or \qquad \displaystyle \lim_{x\rightarrow -\infty}\; f(x)=-4\\ a=-4\qquad or \qquad a=-4\\ so\;\;a=-4$$

so we have

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

Now sub in  ( 1,0) to get the value of b

you will get b=4

$$f(x) = \frac{-4x+4}{x-3}$$

If you think about the asymptotes on a graph it might help. It did help me.

And here is the actual graph

Melody  Aug 24, 2018
edited by Melody  Aug 24, 2018