Afunction $f$ 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 $f(x).$

1. has a horizontal asymptote at

This means that

(for at least one of these limits)

2. has a vertical asymptote at

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

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

This means .

(a) With

the second point above suggests . The first point tells us that

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 so that the linear terms vanish.

The third point tells us that

So

(b) Since

we find that , and and .

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

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