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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 $f(x).$

User Keysha
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1 Answer

3 votes

1.
f has a horizontal asymptote at
y=-4

This means that


\displaystyle\lim_(x\to\pm\infty)f(x)-(-4)=0

(for at least one of these limits)

2.
f has a vertical asymptote at
x=3

This means that
f has a non-removable discontinuity at
x=3. Since
f is some rational function, there must be a factor of
x-3 in its denominator.

3.
f has an
x-intercept at (1, 0)

This means
f(1)=0.

(a) With


f(x)=(ax+b)/(x+c)

the second point above suggests
c=-3. The first point tells us that


\displaystyle\lim_(x\to\pm\infty)(ax+b)/(x-3)+4=0=\lim_(x\to\pm\infty)(ax+b+4x-3)/(x-3)=0

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if
a=-4 so that the linear terms vanish.

The third point tells us that


f(1)=(a+b)/(1-3)=0\implies a=-b\implies b=4

So


f(x)=(-4x+4)/(x-3)

(b) Since


f(x)=(rx+s)/(2x+t)=\frac12(rx+s)/(x+\frac t2)

we find that
\frac t2=-3\implies t=-6, and
r=a=-4 and
s=b=4.

User Rmunn
by
6.7k points
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